Solution Manual for Finite Mathematics An Applied Approach ... · Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at ... Solution Manual

Section 1.1 The Cartesian Plane and Graphing 1

Chapter 1 Applications of Linear Functions Exercises Section 1.1 1. The given points are plotted on the graph below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(3,5)

(6,1)

(-2, -1)(-1, -2)

2. The given points are plotted on the graph below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(4,4)

(0,6)

(0,-2)

(-2,0)


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(5,0)(-7,1)

(4,-3)


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(0,6)(-2,5)

(4,2)

(6,0)

(0, 6)


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-1/2,5)

(4.5,2.5)

(-1,-4)

(0,0.5)


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(3,-5)

(3/4,4)

(-3,-2)

(0,5)

Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young

Full file at https://TestbankDirect.eu/


https://TestbankDirect.eu/Solution-Manual-for-Finite-Mathematics-An-Applied-Approach-3rd-Edition-by-Young

Chapter 1 Applications of Linear Functions 2


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(4,0)

(4,1/2)(0,0)

(4,-0.5)


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-3.5,0.5)

(-2,-3)

(0,4.5)

(3,-4)

9. The coordinates are shown next to the appropriate point in the graph below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(-3,3)

(1,2)

(2,-2)

(-2,-3)

(4,0)

10. The coordinates are shown next to the appropriate point in the graph below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(-1,3)

(0,5)

(5,1)

(-4,-3)

(0,-2)

(2,-4)

(-4,2)

11. The data is shown below:

1996 1997 1998 1999 2000

950

1000

1050

Year

U.S. Exports

Dol

lars

(bill

ions

)


30 40 50 6040

50

60

70

80

Females (millions)

Civilian Labor Forces

U.S

. Mal

es (m

illio

ns)








x200 300 400 500 600

10

20

30

40

50

Gestation (days)

Gestation Periods vs Life Span

Life

Spa

n (y

ears

)


400 500 600 700

800

1000

1200

1400

Morning

Daily English Newspapers

Even

ing


1965 1970 1975 1980 1985 1990 199520

25

30

35

40

45

50

Year

U.S. Smokers

% o

f Sm

oker

s (1

8 an

d ol

der)


1989 1990 1991 1992 1993 1994 1995 1996 1997 199856

58

60

62

64

66

68

70

Year

College Enrollment

% o

f U.S

. Hig

h Sc

hool

Gra

duat

es

17. The Data is shown below:

1988 1990 199819941992 1996

30000

32500

37500

35000

Median Household Income

Med

ian

inco

me

18. Recognizing that the equation is a linear equation, we simply need to find two points on the line. So we will create a representative table using any two values of x we wish to use.

x y ( , )x y

0 2(0) 3 3− = − (0, 3)− 32 ( )3

22 3 0− = ( )32 ,0

The graph is shown at the top of the next page.







Plotting the points and connecting them with a smooth curve we see the following graph:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,-3)

(1.5,0)


x y ( , )x y

0 0 4 4− = − ( )0, 4−

4 4 4 0− = ( )4,0 Plotting the points and connecting them with a smooth curve we see the following graph:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(4,0)

(0,-4)


x y ( , )x y

0 ( )2 0 0− = ( )0,0

2 ( )2 2 4− = − ( )2, 4−


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,0)

(2,-4)


x y ( , )x y

1− ( )5 1 5− = − ( )1, 5− −

1 ( )5 1 5= ( )1,5 Plotting the points and connecting them with a smooth curve we see the following graph:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-1,-5)

(1,5)


x y ( , )x y 0 ( )2 0 6 6− = − ( )0, 6−

3 ( )2 3 6 0− = ( )3,0








x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,-6)

(3,0)


x y ( , )x y

0 ( )2 0 7 7− + = ( )0,7

3 ( )2 3 7 1− + = ( )3,1 Plotting the points and connecting them with a smooth curve we see the following graph:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789


x y ( , )x y 0 ( )1

2 0 2 2− + = ( )0,2

4 ( )12 4 2 0− + = ( )4,0


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789


x y ( , )x y

5− ( )5 5 0− + = ( )5,0−

0 ( )0 5 5+ = ( )0,5 Plotting the points and connecting them with a smooth curve we see the following graph:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(-5,0)

(0,5)

26. Because the exponent on the variable is one, this is a linear function. Therefore, we only need to plot two points. Notice the representative table below:

x ( ) 3 1f x x= − ( )( ),x f x

0 ( ) ( )0 3 0 1 1f = − = − ( )0, 1−

2 ( ) ( )2 3 2 1 5f = − = ( )2,5 The graph is shown at the top of the next page.







Plotting the points in the plane and connecting them with a smooth curve we see the following graph:

x

f(x)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,-1)

(2,5)


x ( ) 2f x x= − ( )( ),x f x

1− ( ) ( )1 2 1 2f − = − − = ( )1,2−

2 ( ) ( )2 2 2 4f = − = − ( )2, 4− Plotting the points in the plane and connecting them with a smooth curve we see the following graph:

x

f(x)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789


t ( ) 4f t t= − + ( )( ),t f t

0 ( ) ( )0 0 4 4f = − + = ( )0,4

4 ( ) ( )4 4 4 0f = − + = ( )4,0

Plotting the points in the plane and connecting them with a smooth curve we see the following graph:

t

f(t)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,4)

(4,0)

29. The exponent on the independent variable is two, the graph will be a parabola. We need to plot several points to get the basic shape. To do this we will simply choose more values of t . However, we still create the same representative table below:

t 2( ) 3P t t= − ( )( ),t P t

2− ( ) ( )22 2 3 1P − = − − = ( )2,1−

1− ( ) ( )21 1 3 2P − = − − = − ( )1, 2− −

0 ( ) ( )20 0 3 3P = − = − ( )0, 3−

1 ( ) ( )21 1 3 2P = − = − ( )1, 2−

2 ( ) ( )22 2 3 1P = − = ( )2,1

Plotting the points in the plane, and connecting them we a smooth curve, we graph the parabola.

t

P(t)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-2,1)

(-1,-2)(0,-3)

(1,-2)

(2,1)







30. Once again, looking at the exponent on the independent variable, we see that it is a one. This is a linear equation, and we only need to plot two points. The representative table is as follows:

p ( ) 2 4Q p p= − ( )( ),p Q p

0 ( ) ( )0 2 0 4 4Q = − = − ( )0, 4−

2 ( ) ( )2 2 2 4 0Q = − = ( )2,0 Plotting the appropriate points and connecting the plots with a smooth curve, we get the graph of the line below:

p

Q(p)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,-4)

(2,0)

31. Once again, we have a linear function. The representative table is shown below:

p ( ) 1Z p p= − ( )( ),p Z p

2− ( ) ( )2 2 1 3Z − = − − = − ( )2, 3− −

5 ( ) ( )5 5 1 4Z = − = ( )5,4 Plotting the points and connecting them with a smooth curve, we see the graph of the line below:

p

Z(p)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-2,-3)

(5,4)

For Problems 32–37, I will be using the standard graphing window on the calculator.

32. Entering 2y x= into the graphing calculator, shows the following graph:

33. Entering 2 1y x= + into the graphing calculator, shows the following graph:

34. Entering 2 2y x= − into the graphing calculator, shows the following graph:







35. Entering 6y x= − into the graphing calculator, shows the following graph:

36. Entering 3y x= into the graphing calculator, shows the following graph:

37. Entering ( )31y x= − into the graphing calculator, shows the following graph:

38. The independent variable in this case is time, t . We would believe that time cannot be negative so the minimum value is 0, the time the removal process began. It would also make sense that there could not be a negative amount of pollutant. So ( ) 0f t ≥ . Which means

( )1201 0 120t t− ≥ ⇒ ≤ . So an appropriate domain for

this application would be 0 120t≤ ≤ . Now create a representative table like the one shown at the top of the next column.

t ( )f t ( )( ),t f t

0 ( ) ( )01200 50 1 50f = − = ( )0,50

60 ( ) ( )6012060 50 1 25f = − = ( )60,25

120 ( ) ( )120120120 50 1 0f = − = ( )120,0

Plotting these arbitrary points and connecting them with a smooth curve, we see the graph of the function:

t (days)

f(t) (pollutant in ppm)

0 20 40 60 80 100 120

10

20

30

40

50

60

(0,50)

(60,25)

(120,0) 39. The independent variable in this function is thousands of miles x . This application implies that 0x ≥ . It also would not make sense to have a negative tread depth, so ( ) 0f x ≥ . This implies that ( )401 0 40x x− ≥ ⇒ ≤ . So

an appropriate domain for this function would be 0 40x≤ ≤ . Creating a representative table using appropriate values from the domain we see:

x ( )f x ( )( ),x f x

0 ( ) ( )0400 2 1 2f = − = ( )0,2

20 ( ) ( )204020 2 1 1f = − = ( )20,1

40 ( ) ( )404040 2 1 0f = − = ( )40,0

The graph is shown at the top of the next page.







Plotting the points from the previous page and connecting them with a smooth curve gives us the graph of the function:

x (1000's of miles)

f(x)

0 10 20 30 40 50

1

2

3

4

(tread depth in cm's)

(0,2)

(20,1)

(40,0)

40. The independent variable in this problem is price p . Thus it makes sense that 0p ≥ . Likewise, the number of DVD’s sold must be positive. This means that

225 0 5 5p p− ≥ ⇒ − ≤ ≤ . However since 0p ≥ , we establish the appropriate domain of 0 5p≤ ≤ . Create a representative table using appropriate points from the domain:

p ( )S p ( )( ),p S p

0 ( ) ( )20 25 0 25S = − = ( )0,25

1 ( ) ( )21 25 1 24S = − = ( )1,24

3 ( ) ( )23 25 3 16S = − = ( )3,16

5 ( ) ( )25 25 5 0S = − = ( )5,0

Plotting the points and connecting them with a smooth curve we see the graph of the function:

p (dollars)

S(p) (thousands of DVD's)

0 1 2 3 4 5 6

2468

10121416182022242628

(0,25)(1,24)

(3,16)

(5,0)

41. The independent variable is price p in hundreds of dollars. Since price must be positive, 0p ≥ . It also makes sense to have a positive number of stereos sold. This means

that ( )2

416 0 8 8p p− ≥ ⇒ − ≤ ≤ . However since price

is positive, and appropriate domain would be 0 8p≤ ≤ . Create a representative table:

p ( ) ( )2

430 16 pS p = − ( )( ),p S p

0 ( ) ( )2040 30 16 480S = − = ( )0,480

2 ( ) ( )2242 30 16 450S = − = ( )2,450

6 ( ) ( )2646 30 16 210S = − = ( )6,210

8 ( ) ( )2848 30 16 0S = − = ( )8,0

Plotting the points and connecting them with a smooth curve we get the graph of the function:

p (100's of dollars)

S(p)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

100

200

300

400

500

(2,450)(0,480)

(6,210)

(8,0)

(# of steroes sold each year)

42. The independent variable in this application is the number of inspections during one week x . Since there are only five inspectors the maximum number of inspections in a given week is 25. So an appropriate domain for this function is 0 25x≤ ≤ . Create a representative table:

x ( )C x ( )( ),x C x

0 ( ) ( )0 500 120 0 500C = + = ( )0,500

10 ( ) ( )10 500 120 10 1700C = + = ( )10,1700

25 ( ) ( )25 500 120 25 3500C = + = ( )25,3500 The graph is shown at the top of the next page.







Plotting the points and connecting them with a smooth curve, we get the graph of the function:

x (# of inspections)

C(x) (cost in dollars)

0 5 10 15 20 25

1000

2000

3000

(0,500)

(10,1700)

(25,3500)

43. The independent variable in this application is the number of shifts needed each week x . Since there are eight students available to work one shift a day, there is a maximum number of 40 shifts per week. The appropriate domain for this function is 0 40x≤ ≤ . Create a representative table using appropriate points from the domain.

x ( ) 80 40C x x= + ( )( ),x C x

0 ( ) ( )0 80 40 0 80C = + = ( )0,80

20 ( ) ( )20 80 40 20 880C = + = ( )20,880

40 ( ) ( )40 80 40 40 1680C = + = ( )40,1680 Plotting the points and connecting them with a smooth curve, we get the graph of the function:

x (# of shifts)

C(x) (cost in dollars)

0 5 10 15 20 25 30 35 40 45

500

1000

1500

(0,80)

(20,880)

(40,1680)

Exercises Section 1.2 1.

a. The slope is 1 5 6 3

6 4 2m − − −= = = −

−

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(4,5)

(6,-1)

2.

a. The slope is 1 ( 1) 0 0

2 3 5m − − −= = =

− − −

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-2,-1) (3,-1)

3.

a. The slope is 1 43 3

5 5 0 01

m−

−= = =− −

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-1,5) (1/3,5)






Section 1.2 Equations of Straight Lines 11

4.

a. The slope is 1 0 1 12 0 2 2

m − − −= = =− − −

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-2,-1)

(0,0)

5.

a. The slope is 7 18 8

5.8 3.7 2.1 16.81

m −= = =

−

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(1,5.8)

(7/8,3.7)

6. a. The slope is

123 250 127 1270 9972.9 1.6 1.3 13 13

m − − −= = = = −

−

b. The graph is:

x

y

-8 -6 -4 -2 0 2 4 6 8

-80-60-40-20

20406080

100120140160180200220240260280

(1.6,250)

(2.9,123)

7.

a. The slope is ( )5 2 7

3 3 0m No Slope

− −= = =

−

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(3,5)

(3,-2)

8.

a. The slope is 8 3 5 57 6 1

m −= = =

−

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(6,3)

(7,8)







9.

a. The slope is 3 1 2 13 1 2

m −= = =

−

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(1,1)

(3,3)

10.

a. The slope is ( )2 2 0 0

6 4 10m −= = =

− −

b. The graph is:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-4,2) (6,2)

11. a. To begin with we will use point-slope to find the slope-intercept form of the line. Substituting the appropriate slope and coordinates we see the equation: 5 2( 2)y x− = − − . Now solve for y to get the slope-intercept form.

5 2 4y x− = − + 5 5 2 4 5y x− + = − + +

Slope-intercept Form: 2 9y x= − + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

2 9y x= − + 2 2 9 2x y x x+ = − + + General Form: 2 9x y+ = b. The graph is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(2,5)

12. a. To begin with we will use point-slope to find the slope-intercept form of the line. Substituting the appropriate slope and coordinates we see the equation:

( )( )6 4 4y x− = − − . Now solve for y to get the slope-intercept form:

6 4 22y x− = + 6 6 4 16 6y x− + = − + +

Slope-intercept Form: 4 22y x= + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

4 10y x= − − 4 4 10 4x y x x+ = − − + General Form: 4 10x y+ = − b. The graph is shown at the top of the next page.







x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-10-9-8-7-6-5-4-3-2-1

123456789

10

(-4,6)


( )235.7 0y x− = − .

Now solve for y to get the slope-intercept form.

235.7y x− =

235.7 5.7 5.7y x− + = +

Slope-intercept Form: 2

3 5.7y x= + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

23 5.7y x= +

2 2 23 3 35.7x y x x− + = + −

General Form: 2

3 5.7 20 30 171x y x y− + = ⇒ − + = b. The graph is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,5.7)


0 12( 4)y x− = − − . Now solve for y to get the slope-intercept form.

12 48y x= − + Slope-intercept Form: 12 48y x= − + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

12 48y x= − + 12 12 48 12x y x x+ = − + + General Form: 12 48x y+ = b. The graph is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(4,0)

15. a. To begin, find the slope of the line passing through the

two points. 1 4 5

6 3 3m − − −= =

−. Now use point-slope to

find the slope-intercept form of the line. Substitute the appropriate slope and one of the known coordinates to get the equation:

534 ( 3)y x−− = − .

We solve for y at the top of the next page to put the equation in slope–intercept form.







534 5y x−− = +

534 4 5 4y x−− + = + +


3 9y x−= + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation as shown:

53 9y x−= +

5 5 53 3 39x y x x−+ = + +

General Form: 5

3 9 5 3 27x y x y+ = ⇒ + = b. The graph is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(3,4)

(6,-1)


two points. 12

2 4 43 7

m −= =− −

. Now use point-slope to

find the slope-intercept form of the line. Substitute the appropriate slope and one of the known coordinates to get the equation:

( )4 17 24y x− = − .


4 27 74y x− = −

4 27 74 4 4y x− + = − +


7 7y x= +

Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

2647 7y x= +

264 4 47 7 7 7x y x x− + = + −

General Form: 264

7 7 4 7 26x y x y− + = ⇒ − + = b. The graph is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-3,2) (1/2,4)


two points. 3 1 202 0.3 23

m − −= =− −

. Now use point-slope

to find the slope-intercept form of the line. Substitute the appropriate slope and one of the known coordinates to get the equation shown at the top of the next column:

20231 ( 0.3)y x−− = − .


20 623 231y x−− = +

20 623 231 1 1y x−− + = + +

Slope-intercept Form: 20 29

23 23y x−= + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

20 2923 23y x−= +

20 20 29 2023 23 23 23x y x x−+ = + +

General Form: 20 29

23 23 20 23 29x y x y+ = ⇒ + =







b. The graph is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(-2,3)(0.3,1)


two points. 700 500 20

20 10m −= =

−. Now use point-slope

to find the slope-intercept form of the line. Substitute the appropriate slope and one of the known coordinates to get the equation:

500 20( 10)y x− = − . Now solve for y to get the slope-intercept form.

500 20 200y x− = − 500 500 20 200 500y x− + = − +

Slope-intercept Form: 20 300y x= + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

20 300y x= + 20 20 300 20x y x x− + = + −

General Form: 20 300x y− + = b. The graph is shown at the top of the next column.

x

y

-8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

200

400

600

800

1000

1200

(10,500)

(20,700)


two points. 150 300 25

23 17m −= = −

−. Now use point-slope

to find the slope-intercept form of the line. Substitute the appropriate slope and one of the known coordinates to get the equation:

300 25( 17)y x− = − − . Now solve for y to get the slope-intercept form.

300 25 425y x− = − + 300 300 25 425 300y x− + = − + +

Slope-intercept Form: 25 725y x= − + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

25 725y x= − + 25 25 725 25x y x x+ = − + + General Form: 25 725x y+ =







b. The graph is shown below:

x

y

-8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

200

400

600

800

1000

1200

(17,300) (23,150)

20. a. Since the x-intercept is the point ( )3,0 and the y-

intercept is the point ( )0, 2− , the slope of the line passing

through the two points is ( )0 2 2

3 0 3m

− −= =

−. Now use

the slope-intercept form of the line to find the equation. Substitute the appropriate slope and the y-intercept into the equation:

( ) ( )23 2y x= + − .

Now solve for y to get the slope-intercept form. Slope-intercept Form: 2

3 2y x= − Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

23 2y x= −

2 2 23 3 32x y x x− + = − −

General Form: 2

3 2 2 3 6x y x y− + = − ⇒ − + = − b. The graph is shown at the top of the column.

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,-2)

(3,0)

21. a. Since the x-intercept is the point ( )1.6,0 and the y-

intercept is the point ( )0,4.3 , the slope of the line passing

through the two points is ( )0 4.3 43

1.6 0 16m

− −= =

−. Now

use the slope-intercept form of the line to find the equation. Substitute the appropriate slope and the y-intercept into the equation:

( ) ( )4316 4.3y x−= + .

Now solve for y to get the slope-intercept form. Slope-intercept Form: 43 43

16 10y x−= + Once the slope-intercept form of the line is known, we isolate the constant term to get the general form of the line. Do this by subtracting the x term from both sides of the equation.

43 4316 10y x−= +

43 43 43 4316 16 10 16x y x x−+ = + + General Form: 43 43

16 10 215 80 344x y x y+ = ⇒ + = b. The graph is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(0,4.3)

(1.6,0)







22. The equation of the horizontal line is 6y = − The equations of the vertical line is 3x = 23. The equation of the horizontal line is 4y =

The equation of the vertical line is 12x −=

24. The equation of the horizontal line is 200y = The equation of the vertical line is 5.7x = − 25. The equation of the horizontal line is 8.6y = − The equation of the vertical line is 1.2x = Note: There are many solutions to 26–31. The following are arbitrary choices. 26. Parallel lines have the same slope, so simply change the constant term in the equation to get a parallel equation. Original line: 5y x=

Parallel Lines:

5 205 155 20

y xy xy x

= −= += +

The graph of each equation is shown below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

y=5xy=5x+15

y=5x+20 y=5x-20

27. Parallel lines have the same slope, so simply change the constant term in the equation to get a parallel equation. Original line: 4 7 6x y− =

Parallel Lines:

4 7 564 7 284 7 28

x yx yx y

− = −− = −− =


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

4x-7y=-56

4x-7y=-28

4x-7y=6

4x-7y=28

28. Parallel lines have the same slope, so simply change the constant term in the equation to get a parallel equation. Original line: 3 7y x= − +

Parallel Lines:

3 33 23 5

y xy xy x

= − −= − += − +


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

y=-3x-15

y=-3x-9

y=-3x+7

y=-3x+15

29. Parallel lines have the same slope, so simply change the constant term in the equation to get a parallel equation. Original line: 5x =

Parallel Lines:

84

2

xxx

= −= −=

The graph of each equation is shown at the top of the next page.







x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789x=-8 x=-4 x=2 x=5

30. Parallel lines have the same slope, so simply change the constant term in the equation to get a parallel equation. Original line: 7y =

Parallel Lines:

84

3

yyy

= −= −=


x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

y=7

y=3

y=-4

y=-8

31. Parallel lines have the same slope, so simply change the constant term in the equation to get a parallel equation. Original line: 2 4 3x y− + = −

Parallel Lines:

2 4 252 4 162 4 32

x yx yx y

− + = −− + =− + =

The graph of each equation is shown at the top of the next column.

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789 -2x+4y=32

-2x+4y=16

-2x+4y=-3

-2x+4y=-25

32. Solve 2 6x y− = for y to find the slope of the line.

122 6 3y x y x− = − + ⇒ = − . Therefore the slope of the

line is 11 2m = . Find the desired equation using point-slope

form:

( )125 3y x− = −

Thus the equation parallel to the given line is:

( ) 71 12 2 25 3y x y x− = − ⇒ = +

33. The given line is in slope-intercept form 3 7y x= − −

Therefore the slope of the line is 1 3m = − . Find the desired equation using point-slope form:

( )2 3 1.3y x− = − − Thus the equation parallel to the given line is:

2 3 3.9 3 5.9y x y x− = − + ⇒ = − + 34. Solve 2 4 7x y− = for y to find the slope of the line.

712 44 2 7y x y x− = − + ⇒ = − .

Therefore the slope of the line is 1

1 2m = . Find the desired equation using point-slope form: ( ) ( )1 1

2 23y x− − = − Thus the equation parallel to the given line is:

131 1 12 4 2 43y x y x+ = − ⇒ = −







35. Solve 2 3 6x y+ = for y to find the slope of the line.

233 2 6 2y x y x−= − + ⇒ = + .

Therefore the slope of the line is 2

1 3m −= . Since the point that we have is the y-intercept, find the desired equation using slope-intercept form:

23 3.5y x−= −

Thus the equation parallel to the given line is:

23 3.5y x−= −

36. a. Since the equation is in slope-intercept form, change the coefficient on the x-term to create a line with the same y-intercept.

New equations:

2 24 26 2

y xy xy x

= − += += +

b. Change the y-coefficient of an equation in slope-intercept form to find the equation of line that has the same x-intercept.

New equations:

2 2 22 2 24 2 2

y xy xy x

− = + ⇒= + ⇒= + ⇒ 1 1

2 2

11

y xy xy x

= − −= +

= +


New equations:

2 44

2 4

y xy xy x

= − −= −= −


New equations: 12

42 4

4

y xy xy x

− = − − ⇒= − − ⇒

= − − ⇒

12

42

2 8

y xy xy x

−

= +

= −

= − −

38. First solve the equation for y to put the equation in slope-intercept form:

323 2 4 2x y y x−+ = ⇒ = +

a. Since the equation is now in slope-intercept form, change the coefficient on the x-term to create a line with the same y-intercept.

New equations:

2 22

2 2

y xy xy x

= − += += +


New equations:

323

231

2 2

22 2

2

y xy xy x

−

−

−

− = + ⇒

= + ⇒

= + ⇒

32

34

21

3 4

y xy xy x

−

= −

= +

= − +


512 22 5x y y x− = ⇒ = −

a. Since the equation is now in slope-intercept form, change the coefficient on the x-term to create a line with the same y-intercept.

New equations:

52

5252

2

5

y xy xy x

= − −

= − −

= −


New equations:

512 2

51 12 2 2

51 12 2 2

y xy x

y x

−

− = − ⇒

= − ⇒

= − ⇒

512 2

55

y xy xy x

−= +

= − += −


New equations:

2 0.60.6

3.2 0.6

y xy xy x

= − −= −= −


New equations: 1412

1.3 0.61.3 0.61.3 0.6

y xy xy x

− = − ⇒

= − ⇒

= − ⇒

1.3 0.65.2 2.42.6 1.2

y xy xy x

= − += −= −








767 6 0x y y x+ = ⇒ =

a, b. Notice that this line goes through the origin. That means that the x-intercept and the y-intercept are the same point. Therefore, simply change the x-coefficient to get the new lines with the same x-intercept and y-intercept.

New Equations:

2

2

y xy xy x

= −==

42. Since y m y m xx

∆= ⇒ ∆ = ∆

∆i (provided 0x∆ ≠ ).

The slope of the line is 5m = . So if: a. 1x∆ = then 5 1 5y∆ = =i b. 2x∆ = then 5 2 10y∆ = =i c. 5x∆ = then 5 5 25y∆ = =i


∆= ⇒ ∆ = ∆


The slope of the line is 3m = . So if: a. 1x∆ = then 3 1 3y∆ = =i b. 2x∆ = then 3 2 6y∆ = =i c. 5x∆ = then 3 5 15y∆ = =i


∆= ⇒ ∆ = ∆


The slope of the line is 6m = − . So if: a. 1x∆ = then 6 1 6y∆ = − = −i b. 2x∆ = then 6 2 12y∆ = − = −i c. 5x∆ = then 6 5 30y∆ = − = −i


∆= ⇒ ∆ = ∆


The slope of the line is 2m = − . So if: a. 1x∆ = then 2 1 2y∆ = − = −i b. 2x∆ = then 2 2 4y∆ = − = −i c. 5x∆ = then 2 5 10y∆ = − = −i


∆= ⇒ ∆ = ∆


The slope of the line is 12m −= . So if:

a. 1x∆ = then 1 1

2 21y − −∆ = =i

b. 2x∆ = then 12 2 1y −∆ = = −i

c. 5x∆ = then 512 25y −−∆ = =i


∆= ⇒ ∆ = ∆


The slope of the line is 25m −= . So if:

a. 1x∆ = then 2 2

5 51y − −∆ = =i

b. 2x∆ = then 2 45 52y − −∆ = =i

c. 5x∆ = then 25 5 2y −∆ = = −i


∆= ⇒ ∆ = ∆


The slope of the line is 12m = . So if:

a. 1x∆ = then 1 1

2 21y∆ = =i

b. 2x∆ = then 12 2 1y∆ = =i

c. 5x∆ = then 512 25y∆ = =i

49. a.

( ) ( )( ) ( )5 800 5 6000 10,000

20 800 20 6000 22,000

S

S

= + =

= + =

The paper had 10,000 subscribers after five weeks, and 22,000 subscribers after 20 weeks. b. Yes, the subscriptions are increasing by 800 subscriptions per week (the slope of the linear function). c. The slope is the number of new subscribers per week after the start of the advertising campaign. The y-intercept is the number of subscribers that were already subscribing to the paper when the advertising campaign started.







d.

t (weeks)

S(t)

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19-2

2

4

6

8

10

12

14

16

18

20

22

24

26

Subs

crib

ers (

in 1

000'

s)

e. ( ) 18,000 800 6000 18,000S t t= ⇒ + = Solve the linear equation by isolating the variable. 800 6000 18,000t + = 800 12,000 15t t= ⇒ = It will take 15 weeks for subscriptions to reach 18,000. 50. a. Since 1995 is the year where 0t = , then 2000 corresponds to the year 5t = and 2010 corresponds to the year 15t = . Substituting the appropriate value of t into the function we see: ( ) ( )0 120 0 600 600S = + =

( ) ( )5 120 5 600 1200S = + =

( ) ( )15 120 15 600 2400S = + = Haworth Motorcycle Company had 600,000 dollars worth of sales in 1995, 1,200,000 dollars worth of sales in 2000, and is predicting 2,400,000 dollars worth of sales in 2010. b. Yes, sales are increasing by 120,000 dollars each year (The slope of the sales function). c. The slope is the yearly increase in sales from the previous year. The y-intercept is the total sales in 1995. d. The graph is shown at the top of the next column.

t (Years since 1995)

S(t) (Sales in $1000's)

0 2 4 6 8 10 12 14 16 18

500

1000

1500

2000

2500

e. Three million is 3000 thousands so ( ) 3,000S t = implies that 120 600 3000t + = .

Solve this equation for t : 120 600 3000 120 2400 20t t t+ = ⇒ = ⇒ = The company expects to reach three million dollars in sales in the year 2015. 51. a. Since August 31st corresponds to 0t = , September 15th corresponds to 15t = and September 20th corresponds to

20t = . Substitute the appropriate values of t into the function. ( ) ( )15 2 15 100 70P = − + =

( ) ( )20 2 20 100 60P = − + = The agent predicts that there will be 70,000 grasshoppers per acre on September 15th, and 60,000 grasshoppers per acre on September 20th. b. According to the model, the grasshopper population is decreasing by 2 thousand grasshoppers per acre each day. c. The slope of the graph is the rate in which the population of grasshoppers per acre is decreasing each day. The y-intercept is the estimated population of grasshoppers per acre that were present on August 31st. d. The domain of this function is 0 50t≤ ≤ because for values of t larger than fifty, the estimated population of grasshoppers is negative. The graph of the function on this domain is shown at the top of the next page.







t (days since 8/31)

P(t) (in 1000's per acre)

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75

20

40

60

80

100

Gra

ssho

pper

Pop

ulat

ion

e. Since ( ) 20 2 100 20P t t= ⇒ − + = , solve the linear equation for t . 2 100 20 2 80 40t t t− + = ⇒ = ⇒ = It will take 40 days for the grasshopper population to reach 20,000 per acre. 52. a. Since 1997 corresponds to 0t = , 1999 corresponds to

2t = , 2000 corresponds to 3t = , and 2006 corresponds to 9t = . Substitute the appropriate values of t into the

function. ( ) ( )2 800 2 18,000 16,400P = − + =

( ) ( )3 800 3 18,000 15,600P = − + =

( ) ( )9 800 9 18,000 10,800P = − + = According to the model, in 1999 there were 16,400 parts per million (ppm) of this pollutant. In 2000 there were 15,600 ppm of this pollutant. It is estimated that there will be 10800 ppm of this pollutant in 2006. b. Yes, the pollutant is decreasing each year by 800 parts per million. c. The slope is amount of the pollutant in parts per million that is eliminated from the lake each year. The y-intercept is the amount of the pollutant in parts per million in 1997. d. The appropriate domain for this function is 0 22.5t≤ ≤ . The graph of the function is shown at the top of the next column.

t (Years since 1997)

P(t) (1000's of ppm)

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32-2

2

4

6

8

10

12

14

16

18

20

Wat

er P

ollu

tant

e. Since ( ) 0 800 18,000 0P t t= ⇒ − + = , solve the linear equation for t .

800 18,000 0 800 18,000 22.5t t t− + = ⇒ = ⇒ = This means that the pollution will be completely gone from the lake midway through the year 2019. 53. a. The corresponding rise in temperature will be 5

9 °Celsius

for each increase of1° Fahrenheit. b. ( ) ( )5 160 2

9 9 380 26C = − = . The corresponding Celsius

temperature will be 2326 26.67° ≈ °Celsius.

c. The original equation is ( ) ( )5 160

9 9C F= − . 9 5 160C F= − 9 160 5C F+ = 9 95 532 32C F F C+ = ⇒ = +

d. The corresponding rise in temperature will be 9

5 °

Fahrenheit for each increase of1° Celsius 54. a. The independent variable is the number of miles driven. So the domain must be positive. Moreover, the depth of the tread on the tire must be positive as well. Using this information the proper domain is 0 40,000x≤ ≤ . The graph is shown at the top of the next page.







x (1000's of miles)

T(x)

0 5 10 15 20 25 30 35 40 45 50 55

-0.2

0.2

0.4

0.6

0.8

1

Trea

d D

epth

(in

inch

es)

b. From the graph we see that the he tread depth on a new tire is one-half inch deep (y-intercept) c. From the graph we see that the tread will last 40,000 miles until it is completely eliminated (depth is zero). 55. a. The graph ( )y M x= is shown below:

Arm bone length (cm's) x

M(x)

0 10 20 30 40 50 60 70

50

100

150

200

250

Mal

e he

ight

(in

cm's)

b. Substitute 46x = in to the function.

( )(46) 2.9 46 70.6 204M = + = . If a male measures 46 centimeters in length from elbow to shoulder, then he is estimated to be 204 centimeters tall. c. Now ( ) 180W x = . Set the equation equal to 180 and solve for x . 2.8 71.5 180 2.8 108.5 38.75x x x+ = ⇒ = ⇒ = A female 180 centimeters tall, should measure 38.75 centimeters from elbow to shoulder.

56. a. The graph is shown below:

x (years since 1970)

A(x)

0 5 10 15 20 25

2468

1012141618202224262830

Med

ian

age

of fi

rst-m

arrie

d m

en

b. Substitute 12x = which corresponds to the year 1982 into the function. ( ) ( )12 0.125 12 21.6 23.1A = + = The median age for first-married men in 1982 was 23.1 years old. c. 0.125 21.6 30 .125 8.4 67.2x x x+ = ⇒ = ⇒ =0.125 21.6 40 .125 18.4 147.2x x x+ = ⇒ = ⇒ = The average age of first-married men will reach 30 of age in the year 2037and reach 40 years of age in 2117. This is not a realistic model over a long period of time, because the average age will continue to increase. It would be hard to imagine that the average age of first-married men could be 100 years old (The year 2597 according to this model). 57. a. Substitute 11x = which corresponds to the year 2006 into the function. (11) 0.91(11) 28.96 38.97S = + = In 2006, the company sales should reach 38,970 cases of soft drinks. b. Solve the equation 0.91 28.96 50x + = for x . 0.91 21.04 23.12x x= ⇒ ≈ . Add this to 1995 to get the answer. The company should reach 50,000 cases in sales in the year 2018 if this trend continues. c. The graph is shown at the top of the next page.








S(x)

0 5 10 15 20 25

10

20

30

40

50

Sale

s (in

$10

00's)

58. a. The graph is shown below for the appropriate values of the domain:


V(x)

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

2000

4000

6000

8000

10^410000

Val

ue (D

olla

rs)

b. Substitute 4x = into the function.

(4) 1333(4) 12,666 7334V = − + = The blue book value of the car would be 7,334 dollars. c. ( ) 8000 1333 12,666 8000V x x= ⇒ − + = Solve the equation for x. 1333 4666x = ⇒ 3.5x ≈ . The car will be 3.5 years old before it depreciates in value to 8000 dollars. Exercises Section 1.3 1. For a function to be linear, it must have a constant rate of change. The increase in yogurt outlets in the first year is 10, followed by an increase of 20 outlets in the second year. Thus the rate of change for this relation is increasing each year. This relation is not modeled by a linear function.

2. For a function to be linear, it must have a constant rate of change. The decrease in yogurt outlets in the first year is 10, followed by a decrease of 10 outlets in the second year, and each year after that. Thus the rate of change for this relation is constant each year. This relation is modeled by a linear function. 3. For a function to be linear, it must have a constant rate of change. The increase in yogurt outlets in the first year is 20, followed by an increase of 20 outlets in the second year, and each year after that. Thus the rate of change for this relation is constant each year. This relation is modeled by a linear function. 4. For a function to be linear, it must have a constant rate of change. The decrease in yogurt outlets in the first year is 100

2 50= , followed by a decrease of 502 25= outlets in the

second year. Thus the rate of change for this relation is decreasing each year. This relation is not modeled by a linear function. 5. For a function to be linear, it must have a constant rate of change. The initial number of outlets is 60. The number of outlets does not change after that, giving the relation a constant rate of change of zero each year. This relation is modeled by a linear function. 6. For a function to be linear, it must have a constant rate of change. The increase in yogurt outlets in the first year is 10(1.20) 12= followed by an increase of 12(1.2) 14≈ outlets in the second year. Thus the rate of change for this relation is increasing each year. This relation is not modeled by a linear function. 7. ( ) 3 20C x x= + a. The marginal cost is the additional cost associated with the production of an additional item. For linear functions, marginal cost is the slope of the line. Thus the marginal cost is $3 per item. b. The fixed cost is a cost that is incurred regardless of number of units produced. For linear functions, fixed cost is the constant term (the coefficient without a variable). Thus, the fixed cost is $20. c. To find the total cost of producing the first 20 items, evaluate the cost function at 20x = . This is shown at the top of the next column. ( ) ( )20 3 20 20 80C = + = . Thus the total cost for

producing the first 20 items is $80.






Section 1.3 Linear Modeling 25

d. Average total cost is the total cost of production divided by the number of items produced. Mathematically it is

( ) 3 20C x xx x

+= . Remember that you have to perform

the operations in the numerator before dividing. To find the average cost of producing the first 20 items, evaluate the average cost function at 20x = .

( )3 20 20 80 420 20

AC+

= = = .

So the average cost of producing the first 20 items is $4. Repeat this process for 100 and 200 items.

( )3 100 20 320 3.20100 100

AC+

= = =

( )3 200 20 620 3.10200 200

AC+

= = =

The average cost of producing the first 100 items is $3.20 and the average cost of producing the first 200 items is $3.10. 8. ( ) 50 3000C x x= + a. The marginal cost is the additional cost associated with the production of an additional item. For linear functions, marginal cost is the slope of the line. Thus the marginal cost is $50 per item. b. The fixed cost is a cost that is incurred regardless of number of units produced. For linear functions, fixed cost is the constant term (the coefficient without a variable). Thus, the fixed cost is $3000. c. To find the total cost of producing the first 20 items, evaluate the cost function at 20x = . ( ) ( )20 50 20 3000 4000C = + = .

Thus the total cost for producing the first 20 items is $4000. d. Average total cost is the total cost of production divided by the number of items produced. Mathematically it is

( ) 50 3000C x xx x

+= .

Remember that you have to perform the operations in the numerator before dividing. To find the average cost of producing the first 20 items, evaluate the average cost function at 20x = .

( )50 20 3000 4000 200

20 20AC

+= = = .

So the average cost of producing the first 20 items is $200. Repeat this process for 100 and 200 items.

( )50 100 3000 8000 80100 100

AC+

= = =

( )50 200 3000 13000 65

200 200AC

+= = =

The average cost of producing the first 100 items is $80 and the average cost of producing the first 200 items is $65. 9. ( ) 3.2 1680C x x= + a. The marginal cost is the additional cost associated with the production of an additional item. For linear functions, marginal cost is the slope of the line. Thus the marginal cost is $3.20 per item. b. The fixed cost is a cost that is incurred regardless of number of units produced. For linear functions, fixed cost is the constant term (the coefficient without a variable). Thus, the fixed cost is $1680. c. To find the total cost of producing the first 20 items, evaluate the cost function at 20x = . ( ) ( )20 3.2 20 1680 1744C = + = .

Thus the total cost for producing the first 20 items is $1744. d. Average total cost is the total cost of production divided by the number of items produced. Mathematically it

is( ) 3.2 1680C x xx x

+= . Remember that you have to

perform the operations in the numerator before dividing. To find the average cost of producing the first 20 items, evaluate the average cost function at 20x = .

( )3.2 20 1680 1744 87.20

20 20AC

+= = = .

The average cost of producing the first 20 items is $87.20. The solution is continued on the next page.







Repeat this process for 100 and 200 items.

( )3.2 100 1680 2000 20100 100

AC+

= = =

( )3.2 200 1680 2320 11.60200 200

AC+

= = =

The average cost of producing the first 100 items is $20 and the average cost of producing the first 200 items is $11.60. 10. ( ) 5.8 2000C x x= + a. The marginal cost is the additional cost associated with the production of an additional item. For linear functions, marginal cost is the slope of the line. Thus the marginal cost is $5.80 per item. b. The fixed cost is a cost that is incurred regardless of number of units produced. For linear functions, fixed cost is the constant term (the coefficient without a variable). Thus, the fixed cost is $2000. c. To find the total cost of producing the first 20 items, evaluate the cost function at 20x = . ( ) ( )20 5.8 20 2000 2116C = + = .


( ) 5.8 2000C x xx x

+= .

Remember that you have to perform the operations in the numerator before dividing. To find the average cost of producing 20 items, evaluate the average cost function at

20x = . ( )5.8 20 2000 2116 105.80

20 20AC

+= = = .

So the average cost of producing the first 20 items is $105.80. Repeat this process for 100 and 200 items.

( )50 100 3000 8000 80100 100

AC+

= = =

( )50 200 3000 13000 65200 200

AC+

= = =

The average cost of producing the first 100 items is $80 and the average cost of producing the first 200 items is $65. 11. ( ) 1.6 5000C x x= + a. The marginal cost is the additional cost associated with the production of an additional item. For linear functions, marginal cost is the slope of the line. Thus the marginal cost is $1.60 per item. b. The fixed cost is a cost that is incurred regardless of number of units produced. For linear functions, fixed cost is the constant term (the coefficient without a variable). Thus, the fixed cost is $5000. c. To find the total cost of producing the first 20 items, evaluate the cost function at 20x = . ( ) ( )20 1.6 20 5000 5032C = + = .


( ) 50 3000C x xx x

+= .


( )1.6 20 5000 5032 251.6020 20

AC+

= = = .

So the average cost of producing the first 20 items is $251.60. Repeat this process for 100 and 200 items.

( )1.6 100 5000 5160 51.60100 100

AC+

= = =

( )1.6 200 5000 5320 26.60200 200

AC+

= = =

The average cost of producing the first 100 items is $51.60 and the average cost of producing the first 200 items is $26.60.







12. ( ) 200 10000C x x= + a. The marginal cost is the additional cost associated with the production of an additional item. For linear functions, marginal cost is the slope of the line. Thus the marginal cost is $200 per item. b. The fixed cost is a cost that is incurred regardless of number of units produced. For linear functions, fixed cost is the constant term (the coefficient without a variable). Thus, the fixed cost is $10,000. c. To find the total cost of producing the first 20 items, evaluate the cost function at 20x = . ( ) ( )20 200 20 10,000 14,000C = + = .

Thus the total cost for producing the first 20 items is $14,000. d. Average total cost is the total cost of production divided by the number of items produced. Mathematically it is

( ) 200 10,000C x xx x

+= .


( )200 20 10,000 14,000 70020 20

AC+

= = = .

The average cost of producing the first 20 items is $700. Repeat this process for 100 and 200 items.

( )200 100 10,000 30,000 300100 100

AC+

= = =

( )200 200 10,000 50,000 250200 200

AC+

= = =

The average cost of producing the first 100 items is $300 and the average cost of producing the first 200 items is $250. 13. Graph each of the functions by plotting x along the horizontal axis and ( ) 8 75C x x= + along the vertical axis. You might need to make a table like the one displayed at the top of the next column.

x ( )

1C x

xy = 2y MC= ( )1,x y ( )2,x y

1 ( )8 1 751 83+ = 8 ( )1,83 ( )1,8

5 ( )8 5 755 23+ = 8 ( )5,23 ( )5,8

10 ( )8 10 7510+ =

15.5

8 ( )10,15.5 ( )10,8

The graphs are shown below:

x (number of items)

$'s

0 1 2 3 4 5 6 7 8 9

10

20

30

40

50

60

70

y2 = marginal cost

y1 = average cost per item

Notice that the marginal cost per item stays a constant $8 per item regardless of production. This implies that the additional cost of each item is $8. However, the average cost per item falls as output is increased. This is due to the fixed cost of production. As output continues to increase, the average fixed cost will get closer and closer to $8 per item. 14. Graph each of the functions by plotting x along the horizontal axis and ( ) 14 125C x x= + along the vertical axis. You might need to make a table like in section 1.1.

x ( )1

C xxy = 2y MC= ( )1,x y ( )2,x y

1 ( )14 1 1251+ =

139

14 ( )1,139 ( )1,14

5 ( )14 5 1255+ =

39

14 ( )5,39 ( )5,14

10 ( )14 10 12510+ =

26.5

14 ( )10,26.5 ( )10,14

The graphs are shown at the top of the next page.







x (number of items)

$'s

0 1 2 3 4 5 6 7 8 9

10

20

30

40

50

60

70

80

90

100

110

120

130

y2 = marginal cost

y1 = average cost per item

Notice that the marginal cost per item stays a constant $14 per item regardless of production. This implies that the additional cost of each item is $14. However, the average cost per item falls as output is increased. This is due to the fixed cost of production. As output continues to increase, the average fixed cost will get closer and closer to $14 per item. 15. Remember that fixed costs are the constant term of a linear cost function that takes the form ( )C x mx b= + .

So 4000b = . Also, it is given that when 20x = , ( )20 10,000C = . Putting these two pieces of

information together we know that ( )20 4000 10,000m + = .

Solve this equation for m by subtracting 4000 from each side of the equation then dividing each side by 20. 20 4000 10,000 20 6000 300m m m+ = ⇒ = ⇒ = Substitute the values of 20m = and 4000b = into the general form of the cost function to get the solution. ( ) 300 4000C x x= + .

16. Remember that fixed costs are the constant term of a linear cost function that takes the form ( )C x mx b= + .

Thus, 3500b = . Also, it is given that when

22x = , ( )22 5000C = . Putting these two pieces of information together we know that ( )22 3500 5000m + = . Solve this equation for m by subtracting 3500 from each side of the equation then dividing each side of the equation by 22. 22 3500 5000 22 1500 68.18m m m+ = ⇒ = ⇒ =

Substitute the values of 68.18m = and 3500b = into the general form of the cost function to get the solution. ( ) 68.18 3500C x x= + .

17. Remember that marginal costs are the slope of a linear cost function that takes the form ( )C x mx b= + . Thus,

25m = . Also, it is given that when 40x = ,

( )40 4000C = . Putting these two pieces of information together we know that ( )25 40 4000b+ = . Solve this equation for b by subtracting 1000 from each side of the equation.

( )25 40 4000 1000 4000 3000b b b+ = ⇒ + = ⇒ = Substitute the values of 25m = and 3000b = into the general form of the cost function to get the solution. ( ) 25 3000C x x= + .

18. Remember that marginal costs are the slope of a linear cost function that takes the form ( )C x mx b= + . Thus,

80m = . Also, it is given that when 30x = , ( )30 6000C = .

Putting these two pieces of information together we know that ( )80 30 6000b+ = . Solve this equation for b by subtracting 2400 from each side of the equation.

( )80 30 6000 2400 6000 3600b b b+ = ⇒ + = ⇒ = Substitute the values of 80m = and 3600b = into the general form of the cost function to get the solution. ( ) 80 3600C x x= + .

19. We have two ordered pairs that represent this cost function; the first is ( )10,2000 and the second

is ( )15,2700 . Find the marginal cost by finding the slope

between these two points 2700 2000 140

15 10m −= =

−.

Now find the equation of the cost function using point-slope form of the line.

( )2000 140 10 2000 140 1400y x y x− = − ⇒ − = − The solution is continued on the next page.







Add 2000 to both sides of the equation to get

140 600y x= + . Thus, the cost function that will give total cost for producing x refrigerators is ( ) 140 600C x x= + . 20. We have two ordered pairs that represent this cost function; the first is ( )20,600 and the second


between these two points 900 600 2035 20

m −= =

−.


( )600 20 20 600 20 400y x y x− = − ⇒ − = − Add 600 to both sides of the equation to get: 20 200y x= + . Thus, the cost function that will give total cost for producing x radios is ( ) 20 200C x x= + . 21. We have two ordered pairs that represent this cost function; the first is ( )50,1000 and the second



60 50m −= =

−.


( )1000 20 50 1000 20 1000y x y x− = − ⇒ − = − Add 1000 to both sides of the equation to get

20 0y x= + . Thus the cost function that will give total cost for producing x ladies’ coats is ( ) 20C x x= . 22. We have two ordered pairs that represent this cost function; the first is ( )12,1400 and the second


between these two points 2250 1400 106.25

20 12m −= =

−.


( )1400 106.25 12y x− = − ⇒

1400 106.25 1275y x− = − . Add 1400 to both sides of the equation to get

106.25 125y x= + . Thus, the cost function that will give total cost for producing x microwave ovens is ( ) 106.25 125C x x= + .

23. We have two ordered pairs that represent this cost function; the first is ( )15,900 and the second



30 15m −= =

−.


( )900 44 15 900 44 660y x y x− = − ⇒ − = − Add 900 to both sides of the equation to get

44 240y x= + . Thus, the cost function that will give total cost for producing x sweepers is ( ) 44 240C x x= + . 24. Since each person uses 192 gallons of water per day, the slope of the linear function that models water usage as a function of people is 192m = . If there aren’t any people 0x = , then there is no water usage, so the y-intercept must be zero. Therefore the linear function that models the number of gallons y used by x people is 192y x= . 25. Since 25% of people skip breakfast each day, the slope of the linear function that models the number of people y who skip breakfast from a group of x people is

.25m = . Since 25% of zero people is still zero, the y-intercept is zero. Therefore, the linear function that models the number of people y who skip breakfast from a group of x people is .25y x= .







26. a. To find a linear function it is necessary to have two points. Let x equal the number of days after January 1. If the store is selling cameras at a rate of four per day, then when 1x = , 150 4 146y = − = . When

2x = , 146 4 142y = − = . Now we have the two

ordered pairs ( )1,146 and ( )2,142 . Find the slope of the

line between these two points 146 142 4

1 2m −= = −

−, then

use point-slope form to find the equation.

( )146 4 1 146 4 4y x y x− = − − ⇒ − = − + Add 146 to both sides to get the linear function that models the number of cameras in the store after January 1.

4 150y x= − + . b. To graph the function plot the y -intercept and then use the slope to decrease the value of y four units for every one unit increase in the value of x . The graph is shown below:

x (number of days)

y

0 5 10 15 20 25 30 35 40 45 50 55-10

102030405060708090

100110120130140150160170

Num

ber o

f cam

eras

in st

ock

c. The slope of the linear function represents decrease of four cameras in the total stock of cameras each day. The y -intercept of the function is the value of the function

when 0x = . Thus the initial stock of 150 cameras is the y -intercept. d. To find out the number of days to reduce the stock of cameras to 90y = cameras. Substitute 90 in for y into the function and solve for x . 90 4 150x= − + Subtract 150 to both sides of the equation and then divide each side by negative four. 90 4 150 60 4 15x x x= − + ⇒ − = − ⇒ = It will take 15 days to reduce the stock of cameras in the store to 90.

27. a. To model a linear function it is necessary to have two points that represent the function. Let x = the number of years since 2000. Since the enrollment is increasing at 600 students per year and the initial population of students is 12,000, when 1x = , 12,000 600 12,600y = + = and when 2x = , 12,600 600 13,200y = + = . So the two ordered pairs that are needed are ( )1,12,600 and ( )2,13,200 . Use these two points to find the slope of the linear function

13,200 12,600 6002 1

m −= =

−.

Now use the point-slope formula to find the equation.

( )12,600 600 1y x− = − ⇒

12,600 600 600y x− = − Add 12,600 to both sides of the equation to get the linear function that models the number of students y enrolled in the college x years after 2000.

600 12,000y x= + . b. To graph the function plot the y -intercept and then use the slope to increase the value of y 600 units for every one unit increase in the value of x . The graph is shown below:

x (years after 2000)

y

-2 0 2 4 6 8 10 12 14 16 18

5000

10^4

1.5x10^4

2x10^4

2.5x10^4 25000

20000

15000

10000

5000

Stud

ent e

nrol

lmen

t

c. The slope of the linear function represents the increase in the number of students each year since 2000. The y -intercept of the function is the value of the function when

0x = . Thus the student enrollment in the year 2000 of 12,000 students is the y -intercept.







d. To find out the number of years it will take for enrollment to reach 20,000y = . Substitute 20,000 for y into the function and solve for x . 20,000 600 12,000x= + Subtract 12,000 from both sides of the equation and then divide each side by 600. 20,000 600 12,000x= + ⇒ 8000 600x= ⇒

13.33x = . Since the enrollment figures are taken on a yearly basis, this means that it will take 14 years for enrollment to reach 20,000 students or the year 2014. 28. a. Let x equal the city’s population and let y = the advertising revenue for the paper. Since sales are increasing at a rate of $250 per each population increase, the slope of the function is 250

1000 .25m = = . Using the point

( )60,000,500,000 and the slope, the point-slope formula becomes

( )500,000 .25 60,000y x− = − . Solve this equation for y

( )500,000 .25 60,000y x− = − ⇒

500,000 .25 15,000y x− = − ⇒ .25 485,000y x= +

This is the linear function that models advertising revenue y as a function of the city’s population x .

b. The slope is the increase in revenue per person. In other words advertising revenue is expected to increase 25 cents per person. The y -intercept is the amount of advertising revenue generated if the city’s population was zero. This is not a realistic number. c. To find the increase in the advertising revenue, multiply the slope by the increase in population ( ) ( ).25 3500 875= . Add the increase in revenue to the revenue that existed in 2000 to get the new amount of advertising revenue of $500,875 .

29. a. Using the methods done in previous problems we notice that the function decreases in value $3000 for each year, hence 3000m = − and the y-intercept of the function is $25,000. Using slope-intercept the linear function is 3000 25,000y x= − + . Where the variable y is the value of the car, and the variable x is the number of years after the initial purchase. b. The slope is the rate of depreciation of the value of the car after it’s initial purchase. The y-intercept is the initial value of the car (the purchase price). c. Substitute the value 5000y = into the function and solve for x to get the number of years it will take for the car to depreciate to $5000. 5000 3000 25,000x= − + ⇒

20000 3000x− = − ⇒ 6.67x =

The value of the car will be $5000 approximately 6.67 years after it is purchased. Likewise to find when the car will be worth $1500, Substitute 1500y = into the function and sole for x . 1500 3000 25,000x= − + ⇒

23,500 3000x− = − ⇒ 7.83x =

The value of the car depreciates to $1500 approximately 7.83 years after it was purchased. d. Graph the function by plotting the y-intercept and using slope depreciate the value 3000 dollars for each year after the car was purchased. The graph is shown below:

x (years after purchase)

y

-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

5000

10^4

1.5x10^4

2x10^4

2.5x10^4 25000

20000

15000

5000

10000

Val

ue (i

n do

llars

)







30. a. The decline in voters is 500 per year, this means that slope is 500m = − . The initial year 2000 had a voter turn out of 22,000, so the y-intercept is 22,000. Using slope-slope intercept formula we see the function is

500 22,000y = − + where y is the voter turnout and x is number of years since 2000. b. Substitute 6x = into the function to get

( )500 6 22,000 19,000y = − + = . There were 19,000 voters in the election of 2006. c. Substitute 12,000y = into the function and solve for x to get the number of years it will take to reach a voter turnout of 12,000. 12,000 500 22,000x= − + ⇒

10,000 500x− = − ⇒ 20x = It will take 20 years for the voter turnout to decline to 12,000 voters. 31. a. Marginal cost is $20 dollars per radio and fixed cost is $2000, using slope-intercept formula the cost function is

20 2000y x= + where y is the total cost of producing x clock radios. The graph is shown below:

x (# of radios)

y (cost in dollars)

0 50 100 150 200 250 300 350 400 450

500100015002000250030003500400045005000550060006500700075008000850090009500

b. Substitute the value 150x = into the function to get

( )20 150 2000 5000y = + = . The total cost of producing 150 radios is $5000. c. The marginal cost of a linear cost function is the slope of the cost function. Since slope is constant over the entire line, then the marginal cost of producing the 201st radio is $20.

d. Find the total cost of producing 200 radios and divide that by 200. So average cost of producing the first 200

radios is ( )20 200 2000

30200

+= .

The average cost of producing 200 radios is 30 dollars per radio. e. Proceeding in the same method of part d, substitute 500, 1000, and 2000 respectively into the equation. The results are shown below:

( )20 500 200024

500+

=

( )20 1000 200022

1000+

=

( )20 2000 200021

2000+

=

The average cost of producing 500 radios is $24 dollars per radio. The average cost of producing 1000 radios is $22 dollars per radio. The average cost of producing 2000 radios is $21 dollars per radio. 32. a. Marginal cost is $80 dollars for each fax and fixed cost is $75,000, using slope-intercept formula the cost function is

80 75000y x= + where y is the total cost of producing x fax/modems. The graph is shown below:

x (# of fax/modems)

y

0 50 100 150 200 250 300 350 400 450-10

102030405060708090

100110120130140150

Cos

t (in

$10

00's)

b. Substitute the value 150x = into the function to get

( )80 150 75000 87000y = + = . The total cost of producing 150 fax/modems is $87,000. c. The marginal cost of a linear cost function is the slope of the cost function. Since slope is constant over the entire line, then the marginal cost of producing the 151st fax/modem is $80.







d. Find the total cost of producing 300 fax/modems and divide that by 300. So average cost of producing the first

300 fax/modems is ( )80 300 75000

330300+

= .

The average cost of producing 300 fax/modems is $330 dollars per fax/modem. Proceeding the same way, substitute 500 into the equation to get:

( )80 500 75000230

500+

=

The average cost of producing 500 fax/modems is $230 dollars per fax/modem. 33. a. The marginal cost of each inspection is $1,500 per year and the fixed cost of monitoring is $25,000. Using slope-intercept formula, the linear function is

1500 25000y x= + where y is the total cost of treatment, and x is the number of inspections. b. Substitute 50x = into the function to get:

( )1500 50 25000 100000y = + = . The total cost of 50 inspections is $100,000. c. The additional cost of the 51st inspection is the marginal cost of $1,500. d. Find the total cost of 50 inspections and divide that by 50. So average cost of 50 inspections is:

( )1500 50 250002000

50+

= .

The average cost of 50 inspections is $2,000. Likewise for 100 and 200 on site inspections

( )1500 100 250001750

100+

=

( )1500 200 250001625

200+

=

The average cost of 100 on-site inspections is $1,750 and the average cost of 200 on-site inspections is $1,625. e. The graph of the function is shown at the top of the next column.

x (# of inspections)

y

0 50 100 150-10

102030405060708090100110120130140150160170180190200

Cos

t (in

$10

00's)

34. a. This means that for each additional dollar you earn over $17,850 you will owe an additional $0.28 in income tax. b. The tax owed on $25,000 dollars will be computed as follows

( ) ( ).15 17,850 .28 25,000 17,850 4679.5T = + − = The total income tax owed on $25,000 is $4,679.50. 35. a. Fixed monthly income will be $1500 per month. This is the amount if you make zero dollars in sales and is equivalent to the y -intercept. The slope of this function is 4% of total sales in the month. Using the slope-intercept formula, the function is .04 1500y x= + where y is total income per month and x is dollars in sales. b. The graph of the function is shown below:

x (Monthly sales)

y

0 10^4 2x10^4 3x10^4

500

1000

1500

2000

2500

3000

10000 20000 30000

Mon

tly in

com

e (D

olla

rs)

c. Substitute 20,000x = into the function to get

( ).04 20,000 1500 2300y = + = . Your monthly income would be $2300. d. Because you make 4% of all sales. The additional income that you would earn would be $0.04.







e. Substitute 3000y = into the function and solve for x to find the dollar amount of sales. 3000 .04 1500x= + ⇒ 1500 .04x= ⇒

37,500x = If you make $37,500 in sales you will earn a monthly income of $3000. 36. a. Let x is the percentage of people unemployed. The rate of declining sales is 50 sets per 1% unemployment, so the slope of the function is 50m = − using the point

( )2,400 the point slope formula yields

( )400 50 2y x− = − − . So the function is

50 500y x= − + . b. The graph is shown below:

x (% unemployment)

y

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16-50

50

100

150

200

250

300

350

400

450

500

550

Num

ber o

f TV

sets

37. a. Let y be the number of apartments rented, and let x be the rent per month. The slope of the line that will model

this relationship is 60 80 1

800 700 5m − −= =

−. Use the point-

slope formula to find the linear equation

( )1580 700y x−− = − ⇒ 1

580 140y x−− = + ⇒ 1

5 220y x−= + The linear function that models occupancy as a function of monthly rent is 1

5 220y x−= + . The domain of this

function is 700 1100x≤ ≤ . The graph of the function is shown at the top of the next column.

x (rent per month)

y

0 200 400 600 800 1000 1200 1400 1600-10

10

20

30

40

50

60

70

80

90

Num

ber o

f apa

rtmen

ts re

nted

b. To check the answer, substitute $700 and $800 into the function to see if you get the correct number of apartments rented.

( )15 700 220 80y −= + =

( )15 800 220 60y −= + =

The function is accurate. c. Substitute $760 into the function to get

( )15 760 220 68y −= + = .

At a monthly rent of $760 the predicted occupancy is 68 apartments. 38. a. Let x be the daily charge per car and let y be the number of cars parked each day. The slope of this function

is50 60 10

4 3m −= = −

−. Using the slope and the

point ( )4,50 , the point-slope equation yields

( )50 10 4y x− = − − . Solve this equation for y to get the function that gives the number of cars parked each day as a function of the charge per car.

( )50 10 4y x− = − − ⇒

50 10 40y x− = − + ⇒ 10 90y x= − +

This is the desired function.







b. Substitute 3.75x = into the function to get

( )10 3.75 90 52.5y = − + = . Since you can not park a fraction of a car, 52 cars will be parked at a charge of $3.75 per day. c. Substitute 80y = into the function and solve for x . 80 10 90x= − + ⇒ 10 10x = ⇒

1x = . The daily fee will need to be $1 in order to park 80 cars. d. Substitute 0y = into the function and solve for x . 0 10 90x= − + ⇒ 10 90x = ⇒

9x = . The daily fee will need to be $9 in order to park zero cars. This does not seem to be realistic, depending on the demand for parking in the area. e. If parking were free, we substitute 0x = into the function to get ( )10 0 90 90y = − + = . The attendants would park 90 cars a day if parking were free. Again, it depends on the demand for parking in the area, but this does not seem realistic. 39. a. Let x be the number of years since 2000 and y be sales in millions. The slope for this function is

3.6 2.7 .452000 1998

m −= =

−.

Since 0x = corresponds to the year 2000 the y-intercept is 3.6. Using slope-intercept form, the equation of the function is 0.45 3.6y x= + . b. Substitute 5x = the number of years since 2000 into the function to get ( )0.45 5 3.6 5.85y = + = . The sales will be $5.85 million in 2005. c. Substitute 10y = into the function and solve for x . 10 .45 3.6x= + ⇒ .45 6.4x = ⇒ 14.2x = This tells us that in the year 2015 sales will top $10 million. d. The linear model probably does not give a good approximation to sales figures for years that far away from the current year because of the complexity of financial markets.

40. Let x be the year and let y be the mortality rate (per

thousand live births). Using the points ( )1990,9.2 and

( )1998,7.2 find the slope of the linear relation

7.2 9.2 11998 1990 4

m − −= =

−. Substitute slope and one point

in to the point-slope formula to get ( )1

49.2 1990y x−− = − . Thus the equation that approximates the mortality rate in a given year is 1

4 506.7y x−= + . The graph of the equation is shown below:

x (year)1985 1990 1995 2000 2005 2010 2015 2020 2025 2030 2035

-1

1

2

3

4

5

6

7

8

9

10

11 y

Mor

talit

y ra

te (p

er 1

000

birth

s)

b. The slope of the line gives the change in mortality rates each year. A slope of 1

4− means the mortality rate declines

0.25 deaths per 1000 live births each year. c. Substitute 1996x = into the equation to get

( )14 1996 506.7 7.7y −= + = .

This does not seem to be as accurate as the other models, since the mortality rate in 1995 was 7.6 which is lower than 7.7. In 2005, ( )1

4 2005 506.7 5.45y −= + = The mortality rate is predicted to be 5.45 deaths per 1000 live births. 41. a. Let x be the number of years since 1960 and let y be the amount of paper and paperboard waste in millions of tons. Since the year 1997 corresponds to the value 37x = , use

the points given find the slope 38.5 5.4 .895

37 0m −= =

−

Now the slope-intercept formula yields .895 5.4y x= + . This equation approximates the amount of waste for a given year.







b. The slope 0.895 is the increase in paper waste generated each year since 1960. c. Substitute 36x = into the equation to check the accuracy ( ).895 36 5.4 37.6y = + = . The model approximates that 37.6 million tons of waste was generated in 1996, where the actual data tells us that 38.0 million tons were generated. The model underestimates the amount of waste generated in 1996 by .4 million tons. d. Substitute 38x = into the equation to check the accuracy ( ).895 38 5.4 39.4y = + = . The model approximates that 39.4 million tons of waste was generated in 1996, where the actual data tells us that 38.2 million tons were generated. The model overestimates the amount of waste generated in 1998 by 1.2 million tons. 42. a. Let x be the number of years since 1995 and y be the dental cost per person. Since the year 1998 corresponds to

3x = the two ordered pairs in this problem are

( )0,165 and ( )3,192 the slope of the linear function can be found using the ordered pairs given in the

model192 165 9

3 0m −= =

−. Since we have the y-intercept

use slope-intercept form to find the equation of the line

3 165y x= + . The graph of the equation is shown below:

x (Years since 1995)

y

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19-20

20

40

60

80

100

120

140

160

180

200

220

240

Den

tal c

ost p

er p

erso

n (D

olla

rs)

b. To check the accuracy, substitute 2x = the number of years 1997 is after 1995 into the model

( )3 2 165 171y = + = . The model estimates dental cost to be $171 dollars per person in 1997, while the actual date shows that dental cost were $184 dollars per person. The model underestimates dental cost per person by $13 in 1997.

c. The slope of the model relates the increase in dental cost per person each year. Our model suggests that dental costs per person are increasing $3 each year. 43. a. To find the average growth per year, find the slope using the two endpoints of the data

220.2 151.5 3.8171998 1980

m −= =

−.

The average growth between of waste between 1980 and 1998 is 3.817 million tons per year. b. Let x be the number of years since 1980, and y be the

waste in millions of tons. Therefore the point ( )0,151.5 is the y -intercept. Use slope-intercept form to find the linear function that approximates y , the waste generated in any year beyond 1980: The function is 3.817 151.5y x= + ; 0 x≤ . Substitute 15x = into the function to get

( )3.817 15 151.5 208.8y = + = The model estimates that 208.8 million tons will be generated in 1995 and the data shows 211.4 million tons of waste was actually generated, so the model underestimates the value by 2.6 million tons. c. Substitute 30x = into the function to get ( )3.817 30 151.5 266.0y = + = . The model predicts that 266 million tons will be generated in the year 2010. d. Substitute 300y = into the function to get 300 3.817 151.5x= + . Solve the equation for x . 300 3.817 151.5x= + ⇒ 148.5 3.817x= ⇒ 38.9x = This model predicts that 2019 will be the first year in which the amount of waste generated will be greater than 300 million tons. ( )1980 39 2019+ =







44. a. To find the average growth per year, find the slope using

the two endpoints of the data 4.3 2.2 .105

2000 1980m −= =

−.

The average growth of persons 85 and older between 1980 and 2000 is 0.105 million per year. b. Let x be the number of years since 1980, and y be the number of people over the age of 85 in millions. Therefore the point ( )0,2.2 is the y -intercept. Use slope-intercept formula to find the linear function that approximates y , the number of people above 85 years of age in any year beyond 1980: The function is 0.105 2.2y x= + ; 0 x≤ . To check for accuracy, substitute 10x = for 1990 and substitute 15x = for 1995 into the function to get

( ).105 10 2.2 3.3y = + =

( ).105 15 2.2 3.8y = + = The model estimates that 3.3 million people will be over the age of 85 in 1990 and the data shows 2.7 million people actually were over the age of 85. The model overestimates the number of people by 0.6 million people. In 1995, the model estimates that 3.8 million people were over the age of 85, and the data states that 3.7 million people were over the age of 85. The model overstates the number of people over the age of 85 by 0.1 million people. c. Substitute 6y = into the function to get 6 0.105 2.2x= + . Solve the equation for x . 6 0.105 2.2x= + ⇒ 3.8 0.105x= ⇒ 36.2x = This model predicts that 2017 will be the first year in which the population of people over the age of 85 will be greater than 3 million. ( )1980 37 2017+ = 45. a. The amount to be depreciated each year is $5000. 100,000 5000

20=

Making this value negative gives us the slope of the depreciation function.

b. Use slope-intercept with the initial value as the y-intercept to get 5000 100,000y x= − + Where y is the book value of the building, and x is the number of years since the building was purchased. c. The graph is shown at the top of the next page. Notice the domain is 0 20x≤ ≤ the time it takes the building to depreciate to zero.


y

0 5 10 15 20 25 30 35-10

10

20

30

40

50

60

70

80

90

100

110

Val

ue (i

n $1

000'

s)

46. a. The amount to be depreciated each year is $2500. 12500 2500

5=

Making this value negative gives us the slope of the depreciation function. b. Use slope-intercept with the initial value as the y-intercept to get 2500 12,500y x= − + Where y is the value of the copier, and x is the number of years since the copier was purchased. c. The graph is shown below. Note that the domain is 0 5x≤ ≤ the time it takes the copier to depreciate to zero.


y

0 1 2 3 4 5 6 7 8-1

123456789

1011121314151617181920

Val

ue (i

n $1

000'

s)







47. a. The amount to be depreciated each year is $3500. 80,000 10,000 3500

20−

=

Making this value negative gives us the slope of the depreciation function. b. Use slope-intercept with the initial value as the y-intercept to get

3500 80,000y x= − + , 0 20x≤ ≤ Where y is the book value of the machine, and x is the number of years since purchased. c. The graph is shown below: Notice the domain is 0 20x≤ ≤ the time it takes the machine to depreciate to $10,000.


y

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24-5

5101520253035404550556065707580859095

100

Val

ue (i

n $1

000'

s)

48. a. The amount to be depreciated each year is $7647.06.

160,000 30,000 7647.06

17−

=

Making this value negative gives us the slope of the depreciation function. b. Use slope-intercept with the initial value as the y-intercept to get 7647.06 160,000y x= − + , 0 17x≤ ≤ Where y is the book value of the apartment building, and x is the number of years since purchased. c. The graph is shown at the top of the next column. Notice the domain is 0 17x≤ ≤ .


y

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

102030405060708090

100110120130140150160170180

Val

ue (i

n $1

000'

s)

49. a. The amount to be depreciated each year is $20,000.

550,000 50000 20,000

25−

=

Making this value negative gives us the slope of the depreciation function. b. Use slope-intercept with the initial value as the y-intercept to get 20,000 550,000y x= − + , 0 25x≤ ≤ Where y is the book value of the apartment building, and x is the number of years since purchased. c. The graph is shown below. Notice the domain is 0 25x≤ ≤ .


y

-1 0 1 2 3 4 5 6 7 8 91011121314151617181920212223242526272829

50

100

150

200

250

300

350

400

450

500

550

Val

ue (i

n $1

000'

s)

50. a. The amount to be depreciated each year is $10,312.50. 90,000 7500 10,312.5

8−

=

Making this value negative gives us the slope of the depreciation function.







b. Use slope-intercept with the initial value as the y-intercept to get 10,312.50 90,000y x= − + , 0 8x≤ ≤ Where y is the book value of the computer, and x is the number of years since purchased. c. Substitute 8x = into the function to get

( )10,312.50 8 90,000 7500y = − + = Since this is the salvage value of the computer the model is accurate. d. The graph is shown below. Notice the domain is 0 8x≤ ≤ .


y

0 1 2 3 4 5 6 7 8 9

10

20

30

40

50

60

70

80

90

100

Val

ue (i

n $1

000'

s)

51. a. When 20p = we can find quantity demanded and quantity supplied by substituting into demand and supply functions. ( ) ( )20 6 20 120S = =

( ) ( )20 160 6 20 40D = − = Supply exceeds demand at this price which means there is a surplus of 80 items at the price of $20. b. Substitute ( ) 100D p = into the demand function and solve for p .

( )100 160 6 p= − ⇒

6 60p = ⇒ 10p = A price of $10 will support a demand of 100 items. At this price: ( ) ( )20 6 10 60S = =

( ) ( )20 160 6 10 100D = − =

Demand exceeds supply at this price which means there is a shortage of 40 items at the price of $10. c. The graphs are shown below:

p (Price in dollars)

Q (quantity)

0 10 20 30 40

102030405060708090

100110120130140150160170

Demand

Supply

Shortage Surplus

52. a. When 20p = we can find quantity demanded and quantity supplied by substituting into demand and supply functions. ( ) ( )20 4 20 1 81S = + =

( ) ( )20 148 3 20 88D = − = Demand exceeds supply at this price which means there is a shortage of 7 items at the price of $20. b. Substitute ( ) 100D p = into the demand function and solve for p .

( )100 148 3 p= − ⇒ 3 48p = ⇒ 16p = A price of $16 will support a demand of 100 items. At this price ( ) ( )16 4 16 1 65S = + =

( ) ( )16 148 3 16 100D = − = Demand exceeds supply at this price which means there is a shortage of 91 items at the price of $8. c. The graphs are shown at the top of the next page.







p (Price in dollars)

Q (quantity)

0 10 20 30 40 50 60 70 80

102030405060708090

100110120130140150160170

Shortage Surplus

Supply

Demand

53. a. Find the slope using the two relations given in the

problem 9 12 36 4 2

m − −= =

−. Now point-slope formula

yields:

( )329 6y x−− = − ⇒

1.5 18y x= − + Letting price be x the demand function is

( ) 1.5 18D p p= − + b. Substitute 5.25p = into the demand function

( ) ( )5.25 1.5 5.25 18 10.1D = − + = the demand for soybeans will be 10.1 million bushels at this price. c. Substitute ( ) 7D p = and solve for price.

7 1.5 18p= − + ⇒ 1.5 11p = ⇒ 7.33p = . A price of $7.33 will support a demand of 7 million bushels. d. At a price of zero dollars, the demand for soybeans will be 18 million bushels. This is not realistic. Demand would be much higher if soybeans were free in order to help alleviate food shortages. 54. a. Find the slope using the two relations given in the

problem 5 15 1

60 20 4m − −= =

−. Now point-slope formula

yields:

( )1415 20y x−− = − ⇒

.25 20y x= − +

Letting price be x the demand function is

( ) .25 20D p p= − + b. Substitute 30p = into the demand function

( ) ( )30 .25 30 20 12.5D = − + = the demand for this product will be 12.5 purchases per thousand people. c. Substitute ( ) 0D p = and solve for price.

0 .25 20p= − + ⇒ .25 20p = ⇒ 80p = . A price of $80 will cause the demand for this product to drop to zero. d. At a price of zero dollars, the demand for this product will be 20 purchases per thousand people. Exercises Section 1.4 1. Since the slopes of the two lines are different, the pair of equations will have one point in common. 2. Since the slopes of the two lines are the same and the y-intercepts are the same, these two lines will have all points in common. 3. Since the slopes of the lines are the same and the y-intercepts are different, these two lines will have no points in common. 4. Since the slopes of the lines are the same and the y-intercepts are the same, these two lines will have all points in common. 5. Since the slopes of the lines are different, these two lines will have one point in common. 6. Since the slopes of the lines are the same, and the y-intercepts are different, these two lines will have no points in common. 7. a. Since the slopes of the two lines are different, this system has only one solution. b. 2 3x y+ =

2x y− = Solve the second equation for x

2 2x y x y− = ⇒ = + Now substitute x into the first equation

( )2 2 3y y+ + = The solution is continued on the next page.






Section 1.4 Two Lines: Relating the Geometry to the Equations 41

Solve the equation on the previous page for y 4 2 3y y+ + = 4 3 3y+ = 3 1y = −

13

y −=

Now substitute the value of y back into the original equation to find x .

( )13 2x −− =

13 2x + =

53

x =

The point of intersection will be ( )5 13 3, − .

8. a. Since the slopes of the two lines are different, this system has only one solution. b.

2 4x y+ = 2 2 3x y− = Add the two equations to eliminate y 3 7x = Now divide by 3 to solve for x

73x =

Now substitute the value of x back into the original equation to find y . 73 2 4y+ =

532y =

56y =

The point of intersection will be ( )7 5

3 6, . 9. a. Since the slopes of the two lines are the same, and the y-intercepts are the same, this system has infinitely many solutions. b. 2 3x y+ =

0.5 1.5x y+ = Let x be the parameter and solve either of the equations for y to find the solution. One method would be to subtract 2x from both sides of equation 1.

2 3y x= − +

The solution is: any real numberx = .

2 3y x= − + 10. a. Since the slopes of the two lines are the same and the y-intercepts are the same, this system has infinitely many solutions. b.

3 2x y+ = 2 6 4x y+ = Let y be the parameter and solve either of the equations for x to find the solution. One method would be to subtract 3y from equation 1.

2 3x y= − any real numbery = .

11. a. Since the slopes of the two lines are different, this system has only one solution. b.

2 0x y+ = 0x y− =

Solve equation 2 for x by adding y to both sides. x y= Now substitute the value of x back into the original equation to find y .

2 0y y+ = 0y =

Substitute the value of y back into one of the original

equations ( )0 0x − = The point of intersection will be ( )0,0 . 12. a. Since the slopes of the two lines are different, this system has only one solution. b. 2 0x y− =

0x y+ = Add the two equations together to eliminate y 3 0x = 0x⇒ = .







Now substitute the value of x back into the original equation to find y .

( )0 0y+ =

0y = The point of intersection will be ( )0,0 . 13. a. Since the slopes of the two lines are the same and the y-intercepts are the same, this system has infinitely many solutions. b.

2 0x y+ = 2 4 0x y+ = Let y be the parameter, solve either of the equations for x to find the solution. One method would be to subtract 2y from both sides of equation 1.

2x y= − The solution is: Let y be any real number.

2x y= − 14. a. The slopes of the two lines are equal, but the y-intercepts are different. This means that there are no solutions. b. Since there is no solution, there are no points of intersection. 15. a. The slopes of the two lines are equal, but the y-intercepts are different. This means that there are no solutions. b. Since there is no solution, there are no points of intersection. 16. a. The slopes of the two lines are equal, but the y-intercepts are different. This means that there are no solutions. b. Since there is no solution, there are no points of intersection. 17. a. Since the slopes of the two lines are different, this system has only one solution. b. 0.5 6x =

3 4x y+ =

To solve this system first solve equation 1 for x by multiplying both sides by 2 0.5 6x =

12x = Now substitute the value of x back into equation 2 in order to find y .

( )12 3 4y+ = subtract 12 from both sides

3 8y = − then divide by 3 8

3y −= . The point of intersection is ( )8

312, − . 18. a. Since the slopes of the two lines are different, this system has only one solution. b.

5y = 2 6x y− = Equation 1is already solved for y

5y = Now substitute the value of y back into equation 2 in order to find x . This is shown at the top of the next page.

( )2 5 6x − = add 5 to both sides

2 11x = then divide by 3 112 5.5x = = .

The point of intersection is ( )5.5,5 . 19. a. Since the slopes of the two lines are different, this system has only one solution. b. 0.5 2 3x y+ =

1x y− = To solve this system first solve equation 2 for x by adding y to both sides

1x y= + Now substitute the value of x back into equation 1 and solve for y as we have done on the top of the next page.







( )0.5 1 2 3y y+ + =

.5 0.5 2 3y y+ + = 0.5 2.5 3y+ = 2.5 2.5y = subtract 0.5 from both sides

1y = . Now substitute the value of y back into the original equation to get x .

( )1 1x − = add one to both sides.

2x = The point of intersection is ( )2,1 . 20. a. Since the slopes of the two lines are different, this system has only one solution. b.

0.6 0x y+ = 2 1x y− =

To solve this system first solve equation 2 for x by adding 2y to both sides

2 1x y= + Now substitute the value of x back into equation 1 in order to find y .

( )2 1 0.6 0y y+ + =

2.6 1 0y + = subtract one from both sides 2.6 1y = − divide by 2.6

513y −= .

Now substitute the value of y back into the original equation to get x .

( )5132 1x −− =

1013 1x + = subtract 10

13 from both sides. 3

13x = The point of intersection is ( )3 5

13 13, − . 21. a. Since the slopes of the two lines are different, this system has only one solution. b. 0.2 0x y− =

0.6 0x y+ =

To solve this system first solve equation 2 for x by adding 0.6y to both sides

0.6x y= Now substitute the value of x back into equation 1 in order to find y .

( )0.2 0.6 0y y− =


( )0.6 0 0x − = add one to both sides.

0x = The point of intersection is ( )0,0 . 22. a. Since the slopes of the two lines are different, this system has only one solution. b. 0.4 0.3 6x y− =

0x y+ = To solve this system first solve equation 2 for x by subtracting y from both sides x y= − Now substitute the value of x back into equation 1 in order to find y .

( )0.4 0.3 6y y− − =

0.7 6y− = Divide both sides by 0.7− 607y −=


( )607 0x −+ = add 60

7 from both sides. 607x =

The point of intersection is ( )60 60

7 7, − . 23. a. Since the slopes of the two lines are different, this system has only one solution.







b. 2 3x y− =

2 6x y− = To solve this system first solve equation 1 for x by adding 2y to both sides

2 3x y= + Now substitute the value of x back into equation 2 in order to find y .

( )2 2 3 6y y+ − =

4 6 6y y+ − = Subtract 6 from both sides 3 0y = Divide by 3


( )2 0 3x − =

3x = The point of intersection is ( )3,0 . 24. a. Since the slopes of the two lines are different, this system has only one solution. b. 0.2 0.7 0x y+ =

0x y− = To solve this system first solve equation 2 for x by adding y to both sides x y= Now substitute the value of x back into equation 1 in order to find y .

( )0.2 .7 0y y+ =

0.9 0y = divide both sides by 0.9 0y = .


( )0 0x − = .

0x = The point of intersection is ( )0,0 .

25. To solve the system using a graphing calculator you first must solve each equation for y Equation 1 Equation 2 2 3 150x y+ = 1.8 3.5 167x y+ = 3 150 2y x= − 3.5 167 1.8y x= −

( )150 23

xy

−=

( )167 1.83.5

xy

−=

Notice that the equations are not simplified. The calculator does not need to see pleasant looking equations so there is no need to simplify them here. Type equation 1 into Y1 of your graphing calculator and type equation 2 into Y2 of your graphing calculator and hit the graph button.

Depending on your window you may not have the picture of the intersection, so you must change your window.

Now hit 2nd Trace to get the calculate screen and select 5: intersection.

The solution to the problem is the following:







26. To solve the system using a graphing calculator you first must solve each equation for y Equation 1 Equation 2

3y x= 3 0.1y x= + Notice that the equations are already solved for y . Type equation 1 into Y1 of your graphing calculator and type equation 2 into Y2 of your graphing calculator.

Change the window to look like

Now hit the graph button.

Now hit 2nd Trace to get the calculate screen and select 5: intersection, the screen should display.

This is because the two lines have the same slope and different y-intercepts. There is no solution to this problem. 27. To solve the system using a graphing calculator you first must solve each equation for y Equation 1 Equation 2 0.3 0.4 0.8x y− = 6 8 16x y− =

0.4 0.8 0.3y x− = − 8 16 6y x− = −

( )( )

0.8 0.30.4

xy

−=

−

( )( )

16 68

xy

−=

−

Notice that the equations are not simplified. The calculator does not need to see pleasant looking equations so there is no need to simplify them here. Type equation 1 into Y1 of your graphing calculator and type equation 2 into Y2 of your graphing calculator.


Now hit the graph button

This time the slopes are equal and so is the y-intercept. There are infinitely many solutions. To double check hit 2nd graph to pull up the table of values.

These points are examples of solutions to the problem. 28. To solve the system using a graphing calculator you first must solve each equation for y Equation 1 Equation 2 0.3 0.4 16x y− = 0.5 21x y+ =

0.4 16 0.3y x− = − 0.5 21y x= −

( )( )

16 0.30.4

xy

−=

−

( )210.5

xy

−=

The solution is continued on the next page.







Notice that the equations are not simplified. The calculator does not need to see pleasant looking equations so there is no need to simplify them here. You type equation 1 into Y1 of your graphing calculator and type equation 2 into Y2 of your graphing calculator.


Now hit 2nd Trace to get the calculate screen and select 5: intersection. The solution to the problem is the following

29. a. The break-even quantity is the quantity where revenue equals costs.

( ) ( )R x C x= 50 20 900x x= + subtract 20x from both sides 30 900x = divide by 30

30x = The break-even quantity is 30 items. b. To find the break-even point substitute 30x = into either function. ( )(30) 50 30 1500R = = So the break-even point is ( )30,1500 c. The graph is shown at the top of the next column.

x (quantity)

$ (dollars)

0 5 10 15 20 25 30 35 40 45

500

1000

1500

2000

(30,1500) C(x)

R(x)


( ) ( )R x C x= 0.5 0.3 20x x= + subtract 0.3x from both sides 0.2 20x = divide by 0.8

100x = The break-even quantity is 100 items. b. To find the break-even point substitute 100x = into either function. ( )(100) 0.5 100 50R = = So the break-even point is ( )100,50 c. The graph is shown below:

x (quantity)

$ (dollars)

0 20 40 60 80 100 120 140 160 180

20

40

60C(x)

R(x)

(100,50)



250x = The break-even quantity is 250 items.







b. To find the break-even point substitute 250x = into either function. ( )(250) 20 250 5000R = = So the break-even point is ( )250,5000 c. The graph is shown below:

x (quantity)

$ (dollars)

0 50 100 150 200 250 300 350 400 450

2000

4000

6000C(x)

R(x)

(250,5000)


( ) ( )R x C x= 6 10 2 90x x+ = + subtract 2x from both sides 4 10 90x + = subtract 10 from both sides 4 80x =

20x = The break-even quantity is 20 items. b. To find the break-even point substitute 20x = into either function. ( )(20) 6 20 10 130R = + = So the break-even point is ( )20,130 c. The graph is shown below:

x (quantity)

$ (dollars)

0 5 10 15 20 25 30 35

50

100

150 C(x)

R(x)

(20,130)



70x = The break-even quantity is 70 items. b. To find the break-even point substitute 70x = into either function. ( )(70) 15 70 1050R = = So the break-even point is ( )70,1050 c. The graph is shown below:

x (quantity)

$ (dollars)

0 20 40 60 80 100 120 140

200

400

600

800

1000

1200

1400C(x)

R(x)

(70,1050)


( ) ( )R x C x= 5 12 3 116x x+ = + subtract 3x from both sides 2 12 116x + = Subtract 12 from both sides 2 102x = Divide by 2

51x = The break-even quantity is 51 items. b. To find the break-even point substitute 51x = into either function. ( )(51) 5 51 12 267R = + =

So the break-even point is ( )51,267







c. The graph is shown below:

x (quantity)

$ (dollars)

0 20 40 60 80

100

200

300

C(x)R(x)

(51,267)


( ) ( )R x C x= 3 2 12x x= + subtract 2x from both sides

12x = The break-even quantity is 12 items. b. To find the break-even point substitute 12x = into either function. ( )(12) 3 12 36R = = So the break-even point is ( )12,36 c. The profit function is revenue minus cost so ( ) ( ) ( )P x R x C x= −

( ) ( )3 2 12P x x x= − +

( ) 12P x x= − d. The graphs are shown below:

x (quantity)

$ (dollars)

0 5 10 15 20-10

10

20

30

40

50

60

70

C(x)

R(x)

P(x)

(12,36)

(12,0)

Notice the break-even quantity is the quantity where the profit function is zero. 36. a. The break-even quantity is the quantity where revenue equals costs.


100x = The break-even quantity is 100 items. b. To find the break-even point substitute 100x = into either function.

( )(100) 50 100 5000R = = So the break-even point is ( )100,5000 c. The profit function is revenue minus cost so ( ) ( ) ( )P x R x C x= −

( ) ( )50 30 2000P x x x= − +

( ) 20 2000P x x= − d. The graphs are shown below:

x (quantity)

$ (dollars)

0 20 40 60 80 100 120 140 160 180

-2000

2000

4000

6000 C(x)R(x)

P(x)

(100, 5000)

(100,0)

Notice the break-even quantity is the quantity where the profit function is zero. 37. a. The break-even quantity is the quantity where revenue equals costs. We calculate the break even quantity at the top of the next page.







( ) ( )R x C x= 20 12 490x x= + subtract 12x from both sides 8 490x = Divide by 8

61.25x = The break-even quantity is 61.25 items. b. To find the break-even point substitute 61.25x = into either function. ( )(61.25) 20 61.25 1225R = = So the break-even point is ( )61.25,1225 c. The profit function is revenue minus cost so: ( ) ( ) ( )P x R x C x= −

( ) ( )20 12 490P x x x= − +


x (quantity)

$ (dollars)

0 20 40 60 80 100

-500

500

1000

1500 C(x)

R(x)

P(x)

(61.25, 1225)

(61.25,0)

Notice the break-even quantity is the quantity where the profit function is zero. 38. a. The break-even quantity is the quantity where revenue equals costs.

( ) ( )R x C x= 60 30 1800x x= + subtract 30x from both sides 30 1800x = Divide by 30

60x = The break-even quantity is 60 items. b. To find the break-even point substitute 60x = into either function. ( )(60) 60 60 3600R = =

So the break-even point is ( )60,3600

c. The profit function is revenue minus cost so ( ) ( ) ( )P x R x C x= −

( ) ( )60 30 1800P x x x= − +


x (quantity)

$ (dollars)

0 10 20 30 40 50 60 70 80 90

-1000

1000

2000

3000

4000 C(x)R(x)

P(x)

(60,3600)

(60,0)

Notice the break-even quantity is the quantity where the profit function is zero. 39. a. Revenue is price times quantity so ( ) 10.8R x x= Cost is variable cost plus fixed cost so ( ) 8.36 8000C x x= +

Profit is revenue minus cost so ( ) ( )10.8 8.36 8000P x x x= − +

( ) 2.44 8000P x x= − b. To break-even ( ) 0P x =

2.44 8000 0x − = Add 8000 to both sides 2.44 8000x = Divide by 2.44

3278.6x = To break-even approximately 3279 C.D.’s must be sold each month. c. The graphs are shown at the top of the next page.







x (quantity)

$ (dollars)

0 1000 2000 3000 4000 5000 6000 7000

10^4

2x10^4

3x10^4

4x10^4 C(x)

R(x)

P(x)

(3279,0)

40000

30000

20000

10000

d. Since both quantities are below the break-even quantity, a loss will be incurred by the company. 40. a. The fixed cost in this case is $400 and the variable cost are $7.50 per person so the cost function is ( ) 7.5 400C x x= +

b. Revenue is price times quantity, so the revenue function is ( ) 10R x x= c. To break-even revenue must equal cost. ( ) ( )R x C x=

10 7.5 400x x= + Subtract 7.5x from each side 2.5 400x = Divide by 2.5

160x = The break-even point is 160 people. d. Profit is revenue minus cost. ( ) ( ) ( )P x R x C x= −

( ) ( )10 7.5 400P x x x= − +

( ) 2.5 400P x x= − e. The graphs are shown below:

x (quantity)

$ (dollars)

0 50 100 150 200 250 300 350

500

1000

1500

2000

2500 C(x)R(x)

P(x)

(160,0)

41. a.

( ) ( )15 0.5 ; 12D p p S p p= − = + To find equilibrium price, set demand equal to supply and solve for p . 15 0.5 12p p− = + add 0.5p to both sides 15 1.5 12p= + subtract 12 from both sides 1.5 3p = divide both sides by 1.5

2p = The equilibrium price is 0 $2p = . b. To find equilibrium quantity substitute 0 $2p = into the demand or supply function.

( ) ( )2 15 0.5 2 14D = − =

( )2 2 12 14S = + = c. The graphs are shown below:

p (price)

Q (quantity)

0 5 10 15 20 25 30

5

10

15

20

25

30

35

(2,14)

D(p)

S(p)

42. a.

( ) ( )12.5 0.25 ; 0.33 9.67D p p S p p= − = + To find equilibrium price, set demand equal to supply and solve for p . 12.5 0.25 0.33 9.67p p− = + add 0.25p to both sides 12.5 0.58 9.67p= + subtract 9.67 from both sides 0.58 2.83p = divide both sides by 0.58

4.88p = The equilibrium price is 0 $4.88p = .







b. To find equilibrium quantity substitute 0 $4.88p = into the demand or supply function.

( ) ( )4.88 12.5 0.25 4.88 11.28D = − =

( ) ( )4.88 0.33 4.88 9.67 11.28S = + = The equilibrium quantity is 0 11.28q = . c. The graphs are shown below:

p (price)

Q (quantity)

0 10 20 30 40 50 60

5

10

15

D(p)

S(p)

(4.88,11.28)

43. a.

( ) ( )400 10 ; 180D p p S p p= − = + To find equilibrium price, set demand equal to supply and solve for p . 400 10 180p p− = + add 10 p to both sides 400 11 180p= + subtract 180 from both sides 11 220p = divide both sides by 11


( ) ( )20 400 10 20 200D = − =

( ) ( )20 20 180 200S = + = The equilibrium quantity is 0 200q = . c. The graphs are shown at the top of the next column.

p (price)

Q (quantity)

0 10 20 30 40

50

100

150

200

250

300

350

400

450

D(p)

S(p)(20,200)

44. a.

( ) ( )148 3 ; 4 1D p p S p p= − = + To find equilibrium price, set demand equal to supply and solve for p . 148 3 4 1p p− = + add 3p to both sides 148 7 1p= + subtract one from both sides 7 147p = divide both sides by 7


( ) ( )21 148 3 21 85D = − =

( ) ( )21 4 21 1 85S = + = The equilibrium quantity is 0 85q = . c. The graphs are shown below:

p (price)

Q (quantity)

0 10 20 30 40 50

25

50

75

100

125

150

175

D(p)

S(p)

(21, 85)







45. a.

( ) ( )50 0.5 ; 12D p p S p p= − = − To find equilibrium price, set demand equal to supply and solve for p . 50 0.5 12p p− = − add 0.5p to both sides 50 1.5 12p= − add 12 to both sides 1.5 62p = divide both sides by 1.5

41.33p = The equilibrium price is 0 $41.33p = . b. To find equilibrium quantity substitute 0 $41.33p = into the demand or supply function.

( ) ( )41.33 50 0.5 41.33 29.3D = − =

( ) ( )41.33 41.33 12 29.3S = − = The equilibrium quantity is 0 29.3q = . This means that the equilibrium quantity is 2,930 units of software. 46. ( ) ( )475 0.35 ; 0.75 120D p p S p p= − = − To find equilibrium price, set demand equal to supply and solve for p . 475 0.35 0.75 120p p− = − add 0.35p to both sides 475 1.1 120p= − add 102 to both sides 1.1 595p = divide both sides by 1.1

540.91p = The equilibrium price is 0 $540.91p = . To find equilibrium quantity substitute 0 $540.91p = into the demand or supply function.

( ) ( )540.91 475 0.35 540.91 285.7D = − =

( ) ( )540.91 0.75 540.91 120 285.7S = − = The equilibrium quantity is 0 285.7q = . However, it does not make much sense to demand .7 of a drum so the equilibrium quantity will be 286 kettle drums. The equilibrium point is ( )540.91,286 . 47. a. The fixed cost for the car is $20,000 and the maintenance cost is $0.40 per mile. Thus the cost function is ( ) .4 20,000C x x= + .

b. Since the car was rented for 160 days, the company gets a fixed revenue of ( ) ( )160 50 8000= for the year. The company also gets a variable revenue of $0.50 per mile. So the revenue function is ( ) .5 8000R x x= + . c. To break-even revenue must equal cost. .5 8000 .4 20, 000x x+ = + subtract .4x from both sides 0.1 8000 20,000x + = subtract 8000 from both sides 0.1 12,000x = Divide by 0.1

120,000x = . The car must be driven 120,000 miles to break-even during the first year. d. The graph is shown below:

x (miles)

$ (dollars)

0 4x10^4 8x10^4 1.2x10^5

2x10^4

4x10^4

6x10^4

8x10^4C(x)

R(x)

(120000, 68000)

20000

40000

60000

80000

40000 80000 120000

48. a. The fixed cost associated with the ditch digger is $36,000. The maintenance or variable cost of the ditch digger is $0.75 per hour use. Thus the cost function is ( ) 0.75 36,000C x x= + where x is number of hours

used. b. The ditch digger rents for $5 per hour of use. Since it was rented for 150 days at $200 a day, the base revenue is ( ) ( )150 200 30,000= . Add this to the per hour fee to get the revenue function of ( ) 5 30,000R x x= + where x is number of hours used.

c. To find the break-even point set revenue equal to cost. ( ) ( )R x C x= and solve for x . The calculations are

shown at the top of the next page.







( ) ( )R x C x=

5 30000 0.75 36,000x x+ = + subtract 0.75x 4.25 30,000 36,000x + = subtract 30,000 4.25 6000x = divide by 4.25

1411.76x = The ditch digger will needed to be used for 1,411.76 hours in the first year to break-even. d. The graphs are shown below:

x (# of hours used)

$ (1000's of dollars)

0 500 1000 1500 2000 2500

10

20

30

40

50

C(x)

R(x)(1411.76,37.05)

49. a. The fixed cost associated with producing the soft-drink container is $50,000. The direct cost of the container is $1.50 per container. Thus the cost function is ( ) 1.50 50,000C x x= + where x is number containers

produced. The revenue function is price times quantity. Since the containers are being sold at $2 a bottle, the revenue function is ( ) 2R x x= b. Profit is revenue minus cost so the profit function is ( ) ( )2 1.50 50,000P x x x= − + which yields

( ) 0.5 50,000P x x= − . The graphs of the three functions are shown below:

x (1000's of containers)


0 50 100 150 200 250

-50

50

100

150

200

250

300

350 C(x)

R(x)

P(x)

(100,0)

c. If the sales department is correct in their estimates, then the company will operate at a loss of $20,000 ( ) ( )60,000 0.5 60,000 50,000 20,000P = − = − the

first year if they sell 60,000 containers. d. To find the break-even point set revenue equal to cost. ( ) ( )R x C x= and solve for x

2 1.5 50,000x x= + subtract1.5x .5 50,000x = divide by 0.5

100,000x = The company would have to sell 100,000 containers to break-even. e. For 20,000x = Total cost is

( ) ( )20, 000 1.50 20, 000 50, 000 80, 000C = + = The total cost is $80,000 dollars. To find average cost of producing 20,000 units divide total

cost by the number of units80,000 420,000

=

Average cost is $4 per unit for the first 20,000 units. For 40,000x = Total cost is

( ) ( )40, 000 1.50 40,000 50,000 110, 000C = + = The total cost is $110,000 dollars. To find average cost of producing 40,000 units divide total

cost by the number of units110,000 2.7540,000

=

Average cost is $275 per unit for the first 40,000 units. f. Marginal cost is the additional cost of producing the next additional unit, so ( ) ( )20, 001 20, 000 80, 001.50 80, 000 1.50C C− = − = .

The marginal cost is $1.50 per unit. This is also equal to the slope of the cost curve. We will use this information in subsequent problems. 50. a. The fixed cost associated with producing the calculator is $100,000. The direct cost of the container is $50 per calculator. Thus the cost function is ( ) 50 100,000C x x= + where x is number of calculators

produced. b. The revenue function is price times quantity. Since the calculators are being sold at $75 each, the revenue function is ( ) 75R x x= . The solution is continued on the next page.







Marginal revenue is the additional revenue associated with selling one additional calculator. Since they are selling the calculators for $75, the marginal revenue is $75. c. Profit is revenue minus cost so the profit function is ( ) ( )75 50 100,000P x x x= − + which yields

( ) 25 100,000P x x= − . The graphs of the three functions are shown below:

x (1000's of calculators)


0 5 10 15 20 25-100

100

200

300

400

500

600

700

800

900

1000

1100C(x)

R(x)

P(x)

(4,0) d. To find the break-even point set revenue equal to cost. ( ) ( )R x C x= and solve for x

75 50 100,000x x= + subtract50x 25 100,000x = divide by 25

4000x = The company would have to sell 4,000 calculators to break-even. e. For 25,000x = Total cost is

( ) ( )25,000 50 25, 000 100, 000 1,350, 000C = + = The total cost is $1.35 million dollars. Marginal cost is the additional cost of producing the next additional unit, since each calculator cost $50 to make, the marginal cost of producing the 25,001st calculator is $50 per unit. This is also equal to the slope of the cost curve. f. To find average cost of producing 25,000 units divide

total cost by the number of units 1,350,000 54

25,000=

Average cost is $54 per unit for the first 25,000 units.

51. a. For the computer method of printing, the fixed cost is $300. Since it cost $25 to print 1000 copies, the linear cost function for the computer printing method is ( ) 25 300C x x= + where x is 1000’s of copies printed.

For the offset process, the fixed cost are $500 dollars and Since it cost $20 to print 1000 copies, the linear cost function is ( ) 20 500C x x= + b. The graphs are shown below:

x (1000's of copies)

$ (dollars)

0 20 40 60 80

250

500

750

1000

1250

1500

1750

2000

2250

2500

2750

Offset

Computer

(40, 1300)

c. From the graph we see that the quantity is 40,000 copies where cost for both printing process equal $1300. d. For any number of copies below: 40,000 (hence 30,000) it would be more advantageous to use the computer method. However, for any number of copies above 40,000 (hence 70,000) it would be more advantageous to use the offset method. 52. a. The fixed cost associated with selling cars is $80,000. The direct cost is $23,000 per car. Thus the cost function is ( ) 23,000 80,000C x x= + where x is number cars sold.

The revenue function is price times quantity. Since the cars are being sold at $25,000 each, the revenue function is ( ) 25,000R x x= .

Profit is revenue minus cost so the profit function is ( ) ( )25,000 23,000 80,000P x x x= − + which yields

( ) 2000 80,000P x x= − .







b. To find the break-even point set revenue equal to cost. ( ) ( )R x C x= and solve for x

25,000 23,000 80,000x x= + subtract 23,000x 2000 80,000x = divide by 2000

40x = The company would have to sell 40 cars per month to break-even. c. The graphs of the three functions are shown below:

x (# of cars sold)


0 10 20 30 40 50 60 70 80 90 100 110 120 130 140

250

500

750

1000

1250

1500

1750

2000

P(x)

C(x)

R(x)

(40,1000)

d. The dealer will experience a profit if more that 40 cars are sold. In this case, the dealer will operate at a loss of $20,000 for selling 30 cars, and a profit of $20,000 for selling 50 cars. 53. a. The cost of brand A is simply what you paid for it since there is not service contract the total cost of ownership is ( ) 500C x = . For brand B, the fixed cost is $350. Adding the service contract, there is a variable cost of 35 dollars per year so the total cost of ownership is ( ) 35 350C x x= + b. The graphs are shown below:

x (# of years)

$ (dollars)

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

100

200

300

400

500

600

700

800

900

Brand A

Brand B

(4.29,500)

c. Set the two cost functions equal to each other and solve for x . 500 35 350x= + subtract 350 from both sides 150 35x= divide by 35

4.29x = . It will take approximately 4.29 years for the two brands to have equal costs. d. Brand A cost $500 at the five-year mark, while brand B cost 525 at the five-year mark. Brand A costs less if you own the stereo for five years. 54. a. At Piedmont Bank, the fixed cost per month is $6 and there are no other fees, so the cost function is ( ) 6C x = At Atlantic, the fixed cost per month is $3 and there is a $0.50 charge per ATM transaction x . The cost function will be ( ) 0.5 3C x x= + b. The graphs are shown below:

x (# of Transactions)

$ (dollars)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

123456789

1011121314

Piedmont

Atlantic

(6,6)

c. Set the two cost equal to each other and solve for x . 0.5 3 6 6x x+ = ⇒ = . Therefore it will take 6 ATM transactions for the two accounts to have the same cost. d. If there are fewer than 6 ATM transactions a month, Atlantic Bank is less costly, however if there are more than 6 ATM transactions a month, Piedmont Bank is less costly. 55. Let x be the number of years after 2000. Finding a linear sales function for A-Mart and B-Mart is the first step. A-Mart had a growth of $1.1 million per year since 2000 and the total sales of $36.3 million in 2000 is the y-intercept so the sales function for A-Mart is ( ) 1.1 36.3S x x= + where sales are in millions of dollars. The solution is continued on the next page.







B-Mart had a growth of $2.3 million per year since 2000 and the total sales of $16.7million is the y-intercept so the sales function for B-Mart is ( ) 2.3 16.7S x x= + Setting the sales function equal to each other we see 1.1 36.3 2.3 16.7x x+ = + Solve for x . 36.3 1.2 16.7x= + 19.6 1.2x=

16.3x = B-Mart will overtake A-Mart in 17 years or the year 2017. 56. Let x be the number of years since 2000. The population of Harrisonville can be modeled using the growth rate of 420 persons per year. Starting with a population of 6254 the linear population model of Harrisonville is ( ) 420 6254H x x= + . The population of Smithville can be modeled using the growth rate of 330 persons leaving per year. Starting with a population of 17,213 the linear population model of Smithville is ( ) 17213 330S x x= − . Set the two equations equal to each other and solve for x . 420 6254 17213 330x x+ = − 750 6254 17213x + = 750 10959x =

14.6x = The population will be the same in 15 years or, 2015. (This is assuming that the census was taken at the end of the year in 2000) 57. Let x be the number of years since 2000. The city of Rumford can model the amount of recycled waste paper by using the average increase of two tons per year. Starting with the 80 tons they recycled in 2000 the linear population model of Rumford is ( ) 2 80R x x= + . The city of Galeton can model the amount of recycled waste paper by using the average increase of three tons per year. Starting with the 38 tons they recycled in 2000 the linear population model of Rumford is ( ) 3 38G x x= + . Set the two equations equal to each other and solve for x . 3 38 2 80x x+ = +

38 80x + = 42x =

The two cities will be recycling the same amount in 42 years, or the year 2042.

58. Two of the boundary lines are the x -axis and y -axis. The equations of those lines are 0y = and

0x = respectively. One of the boundary lines (the flatter one) passes through ( )0,6 and ( )12,0 . This line has slope

0 6 112 0 2

m − −= =

−. Using the slope-intercept formula the

equation of this boundary line is 12 6y x−= +

The final boundary line (the steeper one) passes through

( )0,8 and ( )8,0 . This line has slope 0 8 18 0

m −= = −

−.

Using the slope-intercept formula the equation of this boundary line is 8y x= − + . So the four boundary equations are:

0y = , 0x = , 12 6y x−= + , 8y x= − +

We can find the first three corner points by inspection. The corner points that lie on the axes are ( )0,6 , ( )0,0 and

( )8,0 . To find the fourth and final corner point,

subtract 8y x= − + from 12 6y x−= + line to get

120 2x= − ⇒ 4x =

Substitute the value of 4x = back into either of the boundary lines to get ( )1

2 4 6 4y −= + = The final corner

point is ( )4,4 . The four corner points are ( )0,6 , ( )0,0 , ( )8,0 and ( )4,4 59. The horizontal boundary line that passes through ( )0,6 has the equation 6y = . Likewise the horizontal

boundary line that passes through ( )0,2 has the equation

2y = . The vertical boundary line that passes through the

point ( )8,0 has the equation 8x = . The fourth boundary line passes through the points

( ) ( )0,8 ; 6,0 .The slope of this line is 0 8 46 0 3

m − −= =

−.

Using slope-intercept formula the equation of the boundary line is 4

3 8y x−= + The boundary equations are:

6y = , 2y = , 8x = , 43 8y x−= +







We can find two of the corner points by inspection. The corner points that are formed by the intersection of the horizontal and vertical boundary lines are ( )8,2 and ( )8,6 .

To find the last two corner points substitute 6y = into 43 8y x−= + to get 4

36 8x−= + 43 2x⇒ = ⇒ 3

2x = so

the corner point is ( )32 ,6 .

Likewise substitute 2y = into 4

3 8y x−= + to get 4

32 8x−= + 43 6x⇒ = ⇒ 9

2x = so the corner point is

( )92 , 2 .

The four corner points are: ( )3

2 ,6 , ( )92 , 2 , ( )8,2 and ( )8,6 .

60. The horizontal boundary line that passes through ( )0,1 has equation 1y = . The increasing boundary line

that passes through ( )0,0 and ( )6,5 has slope

5 0 56 0 6

m −= =


equation is 56y x= . The decreasing boundary line that

passes through ( )0,9 and ( )7,0 has slope

9 0 90 7 7

m − −= =


equation is 97 9y x−= + .

The boundary equations are:

1y = , 56y x= , 9

7 9y x−= + It is not safe to choose any of the boundary points by inspection, so substitute 1y = into 5

6y x= to get 5 66 51 x x= ⇒ = . The corner point is ( )6

5 ,1 . To find the

second corner point substitute 1y = into 97 9y x−= + to

get 971 9x−= + ⇒ 9

7 8x = ⇒ 569x = . The corner point

is ( )569 ,1 .

To find the final corner point subtract 5

6y x= from 97 9y x−= + to get 89

420 9x−= + .

Solve this equation for x . 8942 9x = ⇒ 378

89x = .

Substitute the value of x back into one of the boundary lines to get ( )5 378 315

6 89 89y = = . The corner point is

( )378 31589 89, .

The three corner points are ( )6

5 ,1 , ( )569 ,1 and ( )378 315

89 89, . 61. The y-axis is one of the boundary lines, and it as the equation 0x = . The increasing boundary line that passes

through ( )2,0− and ( )4,4 has slope ( )4 0 2

4 2 3m −= =

− −.

Using the point-slope formula the equation is

( ) ( )( )230 2y x− = − − . Distribution gives the boundary

line equation of 2 43 3y x= + The decreasing boundary line

that passes through ( )0,6 and ( )4,4 has

slope6 4 10 4 2

m − −= =

−. Using the slope-intercept formula

the equation is 12 6y x−= + .

The equations of the boundary lines are

0x = , 2 43 3y x= + , 1

2 6y x−= + One of the corner points is already labeled. It is ( )4,4

To find the other corner points substitute 0x = into 2 4

3 3y x= + to get ( )2 4 43 3 30y = + = . The corner

point is ( )430, . To find the second corner point substitute

0x = into 12 6y x−= + to get ( )1

2 0 6 6y −= + = . The

corner point is ( )0,6 . The three corner points are, ( )0,6 ( )4

30, and ( )4,4 . 62. The horizontal boundary line that passes through ( )0,6 has equation 6y = . The increasing boundary line

that passes through ( )0,1 and ( )1,0− has slope

( )1 0 1

0 1m −= =

− −. Using the slope-intercept formula the

equation is 1y x= + . The decreasing boundary line that

passes through ( )0,8 and ( )8,0 has slope

8 0 10 8

m −= = −


equation is 8y x= − + .







The three boundary line equations are

6y = , 1y x= + , 8y x= − + It is not safe to choose any of the boundary points by inspection, so substitute 6y = into 1y x= + to get

6 1 5x x= + ⇒ = . The corner point is ( )5,6 . To find

the second corner point substitute 6y = into 8y x= − + to get 6 8x= − + ⇒ 2x = . The corner

point is ( )2,6 . To find the final corner point

subtract 8y x= − + from 1y x= + to get 0 2 7x= − .

Solve this equation for x . 2 7x = ⇒ 72x = .

Substitute the value of x back into one of the boundary lines to get ( )7 9

2 28y = − + = . The corner point is ( )7 92 2, .

The three corner points are ( )2,6 , ( )5,6 and ( )7 9

2 2, . 63. The y-axis is one boundary line and it has the equation

0x = . The increasing boundary line that passes through

( )2,0− and ( )0,1 has slope ( )1 0 1

0 2 2m −= =

− −. Using

slope intercept form the equation of the boundary line is 12 1y x= +

The decreasing boundary line that passes through

( )10,0 and ( )0,6 has slope 6 0 30 10 5

m − −= =

−. Using

slope intercept form the equation of the boundary line is 35 6y x−= + . The decreasing boundary line that passes

through ( )5,0 and ( )0,8 has slope 8 0 80 5 5

m − −= =

−.

Using slope intercept form the equation of the boundary line is 8

5 8y x−= + . The four boundary line equations are

85 8y x−= + , 3

5 6y x−= + , 12 1y x= + , 0x = .

We can choose find two corner points by inspection. The two corner points on the y-axis are ( )0,1 and ( )0,6 To find the next corner points subtract

85 8y x−= + from 3

5 6y x−= + to get

0 2x= − ⇒ 2x = . Substitute the value of x into one of the boundary lines yields ( )3 24

5 52 6y −= + = . The

boundary point is ( )2452, . To find the final corner point

subtract 85 8y x−= + from 1

2 1y x= + to get 21100 7x= − ⇒ 10

3x = . Substitute the value of x into one

of the boundary lines yields ( )10 812 3 31y = + = . The

boundary point is ( )10 83 3, .

The four boundary points are: ( )10 8

3 3, , ( )2452, , ( )0,1 and ( )0,6 .

Exercises Section 1.5 1. Create the table as shown. We have two points.

n x y x*x x*y 1 1 2 1 2 2 3 7 9 21

Sum 4 9 10 23 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is: 2 4 9b m+ = 4 10 23b m+ = To solve this system multiply the first equation by -2 and add it to the second equation to get 2 5m = which implies that 5

2m = . Substitute this value back into one of the original equations to get ( )5

22 4 9b + =

Solve this equation for b . 2 10 9b + = subtract 10 2 1b = − divide by 2

12b −=

Therefore the least squares regression line for these data points is 5 1

2 2y x= − . b. The graph of the function and the data points are shown at the top of the next page.






Section 1.5 Regression and Correlation 59

x

y

-4 -3 -2 -1 0 1 2 3 4 5 6 7

-2

-1

1

2

3

4

5

6

7

8

9

(1,2)

(3,7)

Notice that with two points the least squares regression line gives the exact fit of the line to the data points. 2. Create the table as shown. We have two points.

n x y x*x x*y 1 -1 2 1 -2 2 6 5 36 30

Sum 5 7 37 28 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 2 5 7b m+ = 5 37 28b m+ = To solve this system multiply the first equation by 5

2− and

add it to the second equation to get 24.5 10.5m = which implies that 3


72 5 7b + =

Solve this equation for b . 15

72 7b + = subtract 157

3472b = divide by 2

177b =


7 7y x= + . b. The graph of the function and the data points are shown at the top of the next column.

x

y

-4 -3 -2 -1 0 1 2 3 4 5 6 7

-2

-1

1

2

3

4

5

6

7

8

9

(-1,2)

(6,5)


n x y x*x x*y 1 -1 3 1 -3 2 2 -1 4 -2

Sum 1 2 5 -5 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 2 2b m+ =

5 5b m+ = − To solve this system multiply the first equation by 1

2− and

add it to the second equation to get 92 6m = − which implies that 4

3m −= . Substitute this value back into one of the original equations to get ( )4

32 2b −+ =

Solve this equation for b . ( )4

32 2b −+ = add 43

1032b = divide by 2

53b =

Therefore the least squares regression line for these data points is 54

3 3y x−= + . b. The graph of the function and the data points are shown at the top of the next page.







x

y

-4 -3 -2 -1 0 1 2 3 4 5 6 7

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

(-1,3)

(2,-1)


n x y x*x x*y 1 3 5 9 15 2 4 8 16 32


2− and

add it to the second equation to get 31

2 2m = which implies that 3m = . Substitute this value back into one of the original equations to get ( )2 7 3 13b + =


4b = − Therefore the least squares regression line for these data points is 3 4y x= − . b. The graph of the function and the data points are shown at the top of the next column.

x

y

-1 0 1 2 3 4 5 6 7 8 9

-2

-1

1

2

3

4

5

6

7

8

9

(3,5)

(4,8)

Notice that with two points the least squares regression line gives the exact fit of the line to the data points. 5. Create the table as shown. We have three points.

n x y x*x x*y 1 1 1 1 1 2 2 4 4 8 3 3 8 9 24

Sum 6 13 14 33 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 3 6 13b m+ = 6 14 33b m+ = To solve this system multiply the first equation by 2− and add it to the second equation to get 2 7m = which implies that 7


23 6 13b + =


83b −=


2 3y x= − . b. The graph of the function and the data points are shown at the top of the next page.







x

y

-1 0 1 2 3 4 5

-2

-1

1

2

3

4

5

6

7

8

9

(1,1)

(2,4)

(3,8)

Notice that with three points the least squares regression line does not give the exact fit of the line to the data points. 6. Create the table as shown. We have three points.

n x y x*x x*y 1 -1 2 1 -2 2 1 0 1 0 3 5 -5 25 -25

Sum 5 -3 27 -27 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 3 5 3b m+ = − 5 27 27b m+ = − To solve this system multiply the first equation by 5

3− and

add it to the second equation to get 563 22m = − which implies that 33

28 1.179m −= ≈ − . Substitute this value back into one of the original equations

to get ( )33283 5 3b −+ = −

Solve this equation for b . 3 5.893 3b − = − add 5.893 3 2.893b = divide by 3

.964b = Therefore the least squares regression line for these data points is 1.179 .964y x= − + . b. The graph of the function and the data points are shown at the top of the next column.

x

y

-2 -1 0 1 2 3 4 5

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

(-1,2)(1,0)

(5,-5)

Notice that with three points the least squares regression line does not give the exact fit of the line to the data points. 7. Create the table as shown. We have four points.

n x y x*x x*y 1 1 1 1 1 2 2 3 4 6 3 3 6 9 18 4 4 5 16 20


4− and

add it to the second equation to get 1525m = which implies that 3


24 10 15b + =

Solve this equation for b . 4 15 15b + = subtract 15 4 0b =

0b = Therefore the least squares regression line for these data points is 3

2y x= . b. The graph of the function and the data points are shown at the top of the next column.







x

y

-2 -1 0 1 2 3 4 5 6 7 8 9

-2

-1

1

2

3

4

5

6

7

8

9

(1,1)

(2,3)

(3,6)

(4,5)

8. Create the table as shown. We have four points.

n x y x*x x*y 1 0 0 0 0 2 1 -1 1 -1 3 2 -3 4 -6 4 5 -6 25 -30

Sum 8 -10 30 -37 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 4 8 10b m+ = − 8 30 37b m+ = − To solve this system multiply the first equation by 2− and add it to the second equation to get 14 17m = − which implies that 17

14m −= . Substitute this value back into one of the original equations

to get ( )17144 8 10b −+ = −

Solve this equation for b . 68

74 10b − = − add 687

274b −= divide by 4

114b −=


14 14y x−= − . b. The graph of the function and the data points are shown at the top of the next column.

x

y

-2 -1 0 1 2 3 4 5 6 7 8 9

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

(0,0)

(1, -1)

(2,-3)

(5, -6)

9. a. The scatter diagram and the regression line are plotted below:

x

y

-2 -1 0 1 2 3 4 5 6 7 8 9

-2

-1

1

2

3

4

5

6

7

8

9

10

11

(1,3)

(2,6)

(3,9)

b. Create the table as shown. We have three points.

n x y x*x x*y y*y 1 2 6 4 12 36 2 1 3 1 3 9 3 3 9 9 27 81

Sum 6 18 14 42 126 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 3 6 18b m+ = 6 14 42b m+ = To solve this system multiply the first equation by 2− and add it to the second equation to get 2 6m = which implies that 3m = . Substitute this value back into one of the original equations to get ( )3 6 3 18b + = The solution is continued on the next page.







Solve this equation for b . 3 18 18b + = subtract 18 4 0b = divide by 4

0b = Therefore the least squares regression line for these data points is 3y x= . c. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑

And substitute the values from the table.

( ) ( ) ( )( ) ( ) ( ) ( )2 2

3 42 6 18

3 14 6 3 126 18r

−=

− −

1r = . The correlation coefficient is one. 10. a. The scatter diagram and the regression line are plotted below:

x

y

-2 -1 0 1 2 3 4 5 6 7 8 9

-2

-1

1

2

3

4

5

6

7

8

9

10

11

(2,6)

(4,4)

(5,3)

b. Create the table as shown. We have three points.

n x y x*x x*y y*y 1 2 6 4 12 36 2 4 4 16 16 16 3 5 3 25 15 9

Sum 11 13 45 43 61 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is displayed at the top of the next column.

3 11 13b m+ = 11 45 43b m+ = To solve this system multiply the first equation by 11

3− and

add it to the second equation to get 14 143 3m −= which implies that 1m = − .

Substitute this value back into one of the original equations to get ( )3 11 1 13b + − = Solve this equation for b . 3 11 13b − = add 11 3 24b = divide by 3

8b = Therefore the least squares regression line for these data points is 8y x= − + . c. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑

And substitute the values from the table:

( ) ( ) ( )( ) ( ) ( ) ( )2 2

3 43 11 13

3 45 11 3 61 13r

−=

− −

1r = − . The correlation coefficient is negative one. 11. a. The scatter diagram and the regression line are plotted on the graph below:

x

y

-2 -1 0 1 2 3 4 5 6 7 8 9

-2

-1

1

2

3

4

5

6

7

8

9

10

11

(3,8)

(5,3)

(6,7)

(8,2)

(9,4)







b. Create the table as shown. We have five points.

n x y x*x x*y y*y 1 3 8 9 24 64 2 5 3 25 15 9 3 6 7 36 42 49 4 8 2 64 16 4 5 9 4 81 36 16

Sum 31 24 215 133 142 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 5 31 24b m+ = 31 215 133b m+ = To solve this system multiply the first equation by 31

5− and

add it to the second equation to get 22.8 15.8m = − which implies that 0.693m = − . Substitute this value back into one of the original equations to get ( )5 31 0.693 24b + − = Solve this equation for b . 5 21.482 24b − = add 21.482 5 45.483b = divide by 5

9.097b = Therefore the least squares regression line for these data points is 0.693 9.097y x= − + . c. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑


( ) ( ) ( )( ) ( ) ( ) ( )2 2

5 133 31 24

5 215 31 5 142 24r

−=

− −

0.64r = − The correlation coefficient is 0.64r = − . 12. a. The scatter diagram and the regression line are plotted at the top of the next column.

x

y

-2 -1 0 1 2 3 4 5 6 7 8 9

-2

-1

1

2

3

4

5

6

7

8

9

10

11

(1,1) (4,1)

(2,3)

(6,2)

(5,6)

(7,9)

b. Create the table as shown. We have six points.

n x y x*x x*y y*y 1 1 1 1 1 1 2 4 1 16 4 1 3 2 3 4 6 9 4 6 2 36 12 4 5 5 6 25 30 36 6 7 9 49 63 81


6− and

add it to the second equation to get 28.83 24.33m = which implies that 0.907m = . Substitute this value back into one of the original equations to get ( )6 25 0.907 22b + = Solve this equation for b . 6 22.671 22b + = add 22.671 6 0.671b = − divide by 6

0.112b = − Therefore the least squares regression line for these data points is 0.907 0.112y x= − .







c. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑


( ) ( ) ( )( ) ( ) ( ) ( )2 2

6 116 25 22

6 131 25 6 132 22r

−=

− −

.656r = The correlation coefficient is .656r = . 13. a. The scatter diagram and the regression line are plotted below:

x

y

-2 -1 0 1 2 3 4 5 6 7 8 9 10 11

-2

-1

1

2

3

4

5

6

7

8

9

10

11

(0,4)

(5,0)

(9,5)

(3,7)

(1,1)


n x y x*x x*y y*y 1 0 4 0 0 16 2 5 0 25 0 0 3 9 5 81 45 25 4 3 7 9 21 49 5 1 1 1 1 1


5− and

add it to the second equation to get equation at the top of the next column.

51.2 5.8m = which implies that 0.113m = . Substitute this value back into one of the original equations to get ( )5 18 0.113 17b + = Solve this equation for b . 5 2.039 17b + = subtract 2.039 5 14.961b = divide by 5

2.992b = Therefore the least squares regression line for these data points is 0.113 2.992y x= + . c. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑


( ) ( ) ( )( ) ( ) ( ) ( )2 2

5 67 18 17

5 116 18 5 91 17r

−=

− −

0.141r = The correlation coefficient is 0.141r = . 14. a. The scatter diagram and the regression line are plotted below:

x

y

-2 -1 0 1 2 3 4 5 6 7 8 9 10 11

-2

-1

1

2

3

4

5

6

7

8

9

10

11 (3,10)

(4,2)

(6,8)

(7,3)

(8,5)








n x y x*x x*y y*y 1 3 10 9 30 100 2 4 2 16 8 4 3 6 8 36 48 64 4 7 3 49 21 9 5 8 5 64 40 25


5− and

add it to the second equation to get 17.2 9.8m = − which implies that .570m = − . Substitute this value back into one of the original equations to get ( )5 28 0.570 28b + − = Solve this equation for b . 5 15.96 28b − = add 15.593 5 43.96b = divide by 5

8.792b = Therefore the least squares regression line for these data points is 0.570 8.792y x= − + . c. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑


( ) ( ) ( )( ) ( ) ( ) ( )2 2

5 147 28 28

5 174 28 5 202 28r

−=

− −

.0351r = − The correlation coefficient is .0351r = − . 15. There should be a very high positive correlation between the number of accidents and the number of automobiles on the highway. The denser the traffic, the better chance an accident will occur. The correlation is probably above 0.9.

16. There should be a positive correlation between the heights and the ages of elementary school-age children. Although some children grow faster than others, the correlation should be around 0.7. 17. There should be almost no correlation between the heights of persons and having bank accounts and the amount of money in their bank account. However, since height is positive, and we like to think that most bank accounts are positive as well, the correlation coefficient will probably end up being a between 0.1 and 0.2. 18. There should be a positive correlation between I.Q. of college students and their GPA. However, it is probably not as strong a correlation as one would think. There is always the possibility of someone having “street smarts” and not “book smarts.” There are also other variables that would come into play, such as extracurricular activities, difficulty of schedule, etcetera that affect GPA. This relation probably has a correlation around 0.5. 19. In general there should be a very high negative correlation. Most old cars have depreciated in value over the years. There are a few classics (like a mint condition ’67 Hemi-Barracuda) that might weaken this negative correlation, but it still should be in the -0.9 to -0.8 range. 20. There should be very little correlation between the student enrollment and the number of football games won in 1990. Small colleges tend to play other small colleges. Some small colleges have good coaches and recruiting and when against bigger schools. Some larger schools don’t have strong athletic programs. The correlation should be around zero. 21. a. Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line. The regression line and the scatter plot are shown below:

x (tons of pollutant)

y

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140

5

10

15

20

25

30

35

40

45

50

55

(20,40)

(30,33) (40,30)

(50,26)

(60,20) (70,16)

(# of Fish)








Create a chart like the following:

n x y x*x x*y y*y 1 20 40 400 800 1600 2 30 33 900 990 1089 3 40 30 1600 1200 900 4 50 26 2500 1300 676 5 60 20 3600 1200 400 6 70 16 4900 1120 256

Sum 270 165 13900 6610 4921 The least squares regression line is calculated to be

0.466 48.457y x= − + . b. Using the formula from earlier or a spreadsheet, the correlation coefficient is 0.995r = − . c. The slope is the key to this solution. According to the linear model, we would expect -0.466 thousand black bass or 466 fish for each ton of pollutant that is introduced into the water. d. Substitute 65x = into the least squares regression line to get ( )0.466 65 48.457 18.2y = − + = The population of black bass should be 18,200 if there is 65 tons of pollutants present. e. Substitute 35 into the least squares regression line to get 35 0.466 48.457x= − + Solve this equation for x . First subtract 48.457 from both sides to get 13.457 0.466x− = − now divide by 0.466−

28.9x = . We can expect 28.9 tons of pollution will result in a population of 35,000 fish.. 22. a. Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line. The regression line and the scatter plot are shown below:

36 37 38 39 40 41 42 43 44 45 46 47

135

140

145

150

155

160

165

y (Height in cm's)

x (Arm bone length in cm's)

Create a chart like the following.

n x y x*x x*y y*y 1 44 164.7 1936 7246.8 27126.092 42.3 160 1789.29 6768 25600 3 41.2 157 1697.44 6468.4 24649 4 41.8 158.6 1747.24 6629.48 25153.965 40.2 154.1 1616.04 6194.82 23746.816 43.6 162.8 1900.96 7098.08 26503.847 40.6 155.3 1648.36 6305.18 24118.098 38.1 147.2 1451.61 5608.32 21667.849 42.5 160 1806.25 6800 25600 10 45 167.1 2025 7519.5 27922.41

Sum 419.3 1586.817618.19 66638.58 252088 The least squares regression line is calculated to be

2.812 40.571y x= + . b. Using the formula from earlier or a spreadsheet, the correlation coefficient is 0.997r = . 23. a. Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line. The regression line and the scatter plot are shown below:

95 100 105 110

35

40

45

50

55

60

65y (Reading readiness score)

x (I.Q. score)

b. Create a chart like the one on the following page.







n x y x*x x*y y*y 1 102 45 10404 4590 2025 2 110 50 12100 5500 2500 3 102 48 10404 4896 2304 4 98 38 9604 3724 1444 5 100 40 10000 4000 1600 6 101 45 10201 4545 2025 7 96 42 9216 4032 1764 8 100 45 10000 4500 2025 9 104 39 10816 4056 1521 10 94 38 8836 3572 1444

Sum 1007 430 101581 43415 18652 The least squares regression line is calculated to be

0.647 22.891y x= − . Plot this line on the graph on the previous page. c. Using the formula from earlier or a spreadsheet, the correlation coefficient is 0.675r = . 24. a. Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line. The regression line and the scatter plot are shown below:

45 50 55 60 65 70 75 80 85 90 95 100 105 110 1154550556065707580859095100105110115 y (performance rating)

x (aptitude test score)

b. Create a chart like the following: (Note: chart continues on the next column.)

n x y x*x x*y y*y 1 80 85 6400 6800 7225

2 86 71 7396 6106 5041

3 87 90 7569 7830 8100

4 84 75 7056 6300 5625

5 91 89 8281 8099 7921

6 60 70 3600 4200 4900

7 83 74 6889 6142 5476

8 93 85 8649 7905 7225

9 82 85 6724 6970 7225

10 87 90 7569 7830 8100

11 72 80 5184 5760 6400

12 80 91 6400 7280 8281

13 79 82 6241 6478 6724

14 81 90 6561 7290 8100

15 82 91 6724 7462 8281

16 92 97 8464 8924 9409

17 91 95 8281 8645 9025

18 83 90 6889 7470 8100

19 89 86 7921 7654 7396

20 85 80 7225 6800 6400

Sum 1667 1696 140023 141945 144954 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 20 1667 1696b m+ = 1667 140023 141945b m+ = To solve this system multiply the first equation by 83.35− and add it to the second equation to get 1078.55 583.4m = which implies that 0.541m = . Substitute this value back into one of the original equations to get ( )20 1667 0.541 1696b + = Solve this equation for b . 20 901.7 1696b + = 20 794.3b =

39.715b = The least squares regression line is calculated to be

0.541 39.715y x= + . Where y is the employees performance rating, and x is the employees test result. Plot this line on the graph in the previous column. c. Using the correlation coefficient formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 2

20 141, 945 1667 1696

20 140, 023 1667 20 144, 954 1696r

−=

− −

The correlation coefficient is 0.5277r = .







25. a. Create a table like the one shown below:

n x y x*x x*y y*y 1 17 19 289 323 361 2 19 25 361 475 625 3 21 33 441 693 1089 4 23 57 529 1311 3249 5 25 71 625 1775 5041 6 28 113 784 3164 127697 32 123 1024 3936 151298 38 252 1444 9576 635049 39 260 1521 10140 67600

10 41 293 1681 12013 85849Sum 283 1246 8699 43406 255216

Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 10 283 1246b m+ = 283 8699 43406b m+ = To solve this system multiply the first equation by 28.3− and add it to the second equation to get 690.1 8144.2m = which implies that 11.801m = . Substitute this value back into one of the original equations to get ( )10 283 11.801 1246b + = Solve this equation for b . 10 3339.818 1246b + = 10 2093.818b = − divide by 5

209.382b = − Therefore the least squares regression line for these data points is 11.801 209.382y x= − . where x is the diameter of the tree in inches and y is volume in hundreds of board feet that can be retrieved from each tree. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑

And substitute the values from the table as shown at the top of the next column.

( ) ( ) ( )( ) ( ) ( ) ( )2 2

10 43406 283 1246

10 8699 283 10 255216 1246r

−=

− −

0.981r = The correlation coefficient is 0.981r = . b. The rate of change between board feet and the inches in diameter of the tree is the slope of the regression line. For each additional inch in diameter, the regression model predicts an increase in 11.801 hundred feet of boards, or 1,180.1 additional board feet. c. Substitute 45x = into the regression equation to get

( )11.801 45 209.382 321.663y = − = . The regression model predicts that a tree with a diameter of 45 inches will produce 321.663 hundred feet of boards or 32,166 feet of boards. d. To find out how large a tree was in diameter that produced 250 hundred (25000) board feet, substitute

250y = into the regression equation and solve for x .

( )250 11.801 209.382x= −

( )459.382 11.801 x=

38.93x = . The regression model predicts that the tree that produces 25000 feet of boards is 38.93 inches in diameter. 26. a. Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line. The regression line and the scatter plot are shown below:

x (wages in $1000's)

y

0 5 10 15 20 25 30 35 40 45 50 55 60 65

1

2

3

4

5

6

7

8

9 (Sales volume in hundreds of thousands)







b. Create a chart like the following.

n x y x*x x*y y*y 1 30.2 3.7 912.04 111.74 13.69

2 52.1 4.8 2714.41 250.08 23.04

3 23.6 2.7 556.96 63.72 7.29

4 38.2 3.5 1459.24 133.7 12.25

5 45.8 4.7 2097.64 215.26 22.09

6 59.2 5.1 3504.64 301.92 26.01

7 41.7 3.9 1738.89 162.63 15.21

8 62.1 5.4 3856.41 335.34 29.16

Sum 352.9 33.8 16840.23 1574.39 148.74 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 8 352.9 33.8b m+ = 352.9 16840.23 1574.39b m+ = To solve this system multiply the first equation by

44.1125− and add it to the second equation to get 1272.93 83.39m = which implies that 0.0655m = . Substitute this value back into one of the original equations to get ( )8 352.9 0.0655 33.8b + = Solve this equation for b . 8 23.12 33.8b + = 8 10.68b =


0.0655 1.335y x= + , where y is sales volume in hundreds of thousands of dollars, and x is the employees wage in thousands of dollars. Plot this line on the graph above. Using the correlation coefficient formula

( ) ( ) ( )( ) ( ) ( ) ( )2 2

8 1574.39 352.9 33.8

8 16840.23 352.9 8 148.74 33.8r

−=

− −

The correlation coefficient is 0.959r = . c. The value 7y = corresponds to $700,000. Substitute this value into the regression equation and solve for x . 7 0.0655 1.335x= + 5.665 0.0655x=

86.49x = According to the regression model, wages of $86,490 will be spent to support a sales volume of $700,000.

27. a. Create a table like the ones shown below for each set of data. Current Dollars

n x y x*x x*y y*y 1 0 4.88 0 0 23.81442 5 5.94 25 29.7 35.28363 10 6.75 100 67.5 45.56254 15 7.69 225 115.35 59.13615 19 9.08 361 172.52 82.4464

Sum 49 34.34 711 385.07 246.243 Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation for current dollars to be

0.210 4.81y x= + , where y is average hourly earnings in current dollars and x is the year since 1980. To find the correlation coefficient use the formula and substitute the values from the table:

( ) ( ) ( )( ) ( ) ( ) ( )2 2

5 385.07 49 34.34

5 711 49 5 246.243 34.34r

−=

− −

0.991r = The correlation coefficient between hourly average earnings in current dollars and the year since 1980 is 0.991r = . Constant Dollars

n x y x*x x*y y*y 1 0 5.7 0 0 32.49 2 5 5.39 25 26.95 29.05213 10 5.07 100 50.7 25.70494 15 4.97 225 74.55 24.70095 19 5.39 361 102.41 29.0521

Sum 49 26.52 711 254.61 141 Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation for constant dollars to be

0.0229 5.53y x= − + , where y is average hourly earnings in constant dollars and x is the year since 1980. To find the correlation coefficient use the formula and substitute the values from the table as shown at the top of the next page.







( ) ( ) ( )( ) ( ) ( ) ( )2 2

5 254.61 49 26.52

5 711 49 5 141 26.52r

−=

− −

0.599r = − . The correlation coefficient between hourly average earnings in constant dollars and the year since 1980 is 0.991r = . b. Plot the year since 1980 on the x-axis and then draw the regressions lines on the same coordinate plane. The graph is shown below:

x (year since 1980)

y (Hourly earnings in $)

0 5 10 15 20-0.5

0.51

1.52

2.53

3.54

4.55

5.56

6.57

7.58

8.59

9.5

Constant Dollars

Current Dollars

c. Substitute 25x = into the appropriate regression equation to get Current Dollars: ( )0.210 25 4.81 10.06y = + = Constant Dollars: ( )0.0229 25 5.53 4.96y = − + = Subtract the two numbers to find the difference between current dollar and constant dollar average hourly earnings in 2005. We see that the difference is 10.84 4.96 5.88− = or there is a difference of $5.88 between current dollar and constant dollar average hourly earnings. 28. a. Using techniques done in the first part of this section, or using a spreadsheet we can find the least squares regression line. The regression line and the scatter plot are shown at the top of the next column.

x (Temp degree's F)

y

0 10 20 30 40 50 60 70 80 90 100 110 120 130

10

20

30

40

50

60

70

80

(# of seconds to count 50 chirps)

b. Create a chart like the following.

n x y x*x x*y y*y 1 45 94 2025 4230 8836 2 52 59 2704 3068 3481 3 55 42 3025 2310 1764 4 58 36 3364 2088 1296 5 60 32 3600 1920 1024 6 62 32 3844 1984 1024 7 65 26 4225 1690 676 8 68 24 4624 1632 576 9 70 21 4900 1470 441

10 75 20 5625 1500 400 11 82 17 6724 1394 289 12 83 16 6889 1328 256

Sum 775 419 51549 24614 20063 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 12 775 419b m+ = 775 51549 24614b m+ = To solve this system multiply the first equation by

64.583− and add it to the second equation to get 1497.175 2446.3m = − which implies that

1.6339m = − . Substitute this value back into one of the original equations to get ( )12 775 1.6339 419b + − = Solve this equation forb . 12 1266.64 419b − = 12 1685.64b =

140.47b = The solution is continued on the next page.







The least squares regression line is calculated to be 1.63 140.47y x= − + , where y is number of seconds it

takes to count to 50 chirps, and x is the temperature in degrees Fahrenheit. Plot this line on the graph on the previous page. Using the correlation coefficient formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 2

12 24614 775 419

12 51549 775 12 20063 419r

−=

− −

The correlation coefficient is 0.858r = − . c. Substitute 0x = into the regression equation to get

( )1.63 0 140.47 140.47y = − + = . The model predicts it would take approximate 140.47 seconds to count to 50 chirps when the temperature is zero degrees Fahrenheit. d. Substitute 1y = into the regression equation to get 1 1.63 140.47x= − + and then solve for x . 1.63 139.47x =

85.6x = The model predicts in order to count to 50 chirps in 1 second, the temperature would have to be 85.6 degrees Fahrenheit. 29. a. Create a chart like the following.

n x y x*x x*y y*y 1 0 79 0 0 6241 2 5 135 25 675 18225 3 9 135 81 1215 18225 4 10 163 100 1630 26569 5 12 279 144 3348 77841

Sum 36 791 350 6868 147101 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 5 36 791b m+ = 36 350 6868b m+ = To solve this system multiply the first equation by 7.2− and add it to the second equation to get 90.8 1172.8m = which implies that 12.916m = . Substitute this value back into one of the original equations to get ( )5 36 12.916 791b + =

Solve the equation in the previous column for b . 5 464.976 791b + = 5 326.024b =


12.916 65.2y x= + , where y is number of millions of dollars spent on magazine advertising for drugs and remedies, and x is number of years since 1980. Using the correlation coefficient formula

( ) ( ) ( )( ) ( ) ( ) ( )2 2

5 6868 36 791

5 350 36 5 147101 791r

−=

− −

The correlation coefficient is 0.83r = . b. Substitute 30x = into the regression equation to get

( )12.916 30 65.2 452.68y = + = . According to the regression equation there will be $452.68 million spent on magazine advertising for drugs and remedies. 30. a. Create a chart like the following.

n x y x*x x*y y*y 1 1 5.47 1 5.47 29.92092 2 4.98 4 9.96 24.80043 3 4.17 9 12.51 17.38894 4 3.81 16 15.24 14.51615 5 3.92 25 19.6 15.36646 6 4.05 36 24.3 16.40257 7 3.93 49 27.51 15.44498 8 3.9 64 31.2 15.21 9 9 3.95 81 35.55 15.6025

10 10 3.97 100 39.7 15.7609Sum 55 42.15 385 221.04 180.4135

Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 10 55 42.15b m+ = 55 385 221.04b m+ = To solve this system multiply the first equation by 5.5− and add it to the second equation to get 82.5 10.785m = − which implies that .131m = − . Substitute this value back into one of the original equations to get ( )10 55 .131 42.15b + − =







Solve the equation on the previous page for b . 10 7.19 42.15b − = 10 49.34b =


0.131 4.934y x= − + , where y lead emissions in thousands of metric tons, and x is number of years starting with 1988. Using the correlation coefficient formula

( ) ( ) ( )( ) ( ) ( ) ( )2 2

10 221.04 55 42.15

10 385 55 10 180.4135 42.15r

−=

− −

The correlation coefficient is 0.716r = − . c. Substitute 0y = into the regression equation to get 0 0.131 4.934x= − + . Solve this equation for x . 0.131 4.934x =

37.66x = The regression equation predicts that it will take 37.66 years after 1988, or by the year 2026 lead emissions will be reduced to zero. 31. a. Create a table like the ones shown below for each set of data. Gross waste generated (in millions of tons)

n x y x*x x*y y*y 1 0 87.8 0 0 7708.842 10 121.9 100 1219 14859.613 20 151.5 400 3030 22952.254 25 164.4 625 4110 27027.365 27 178.1 729 4808.7 31719.616 30 205.2 900 6156 42107.047 35 211.4 1225 7399 44689.968 38 220.2 1444 8367.6 48488.04

Sum 185 1340.5 5423 35090.3 239552.7 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 8 185 1340.5b m+ = 185 5423 35090.3b m+ =

Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation gross waste generated to be

3.57 84.925y x= + , where the variable y is the amount of waste generated in millions of tons and x is the year since 1960. To find the correlation coefficient use the formula, and substitute the values from the table.

( ) ( ) ( )( ) ( ) ( ) ( )2 2

8 35181.3 185 1343.1

8 5423 185 8 240658.8 1343.1r

−=

− −

0.989r = .

The correlation coefficient between tons of waste generated and years since 1960 is 0.989r = . Materials recovered (in millions of tons)

n x y x*x x*y y*y 1 0 5.9 0 0 34.81 2 10 8.6 100 86 73.96 3 20 14.5 400 290 210.254 25 16.4 625 410 268.965 27 20.4 729 550.8 416.166 30 33.6 900 1008 1128.967 35 54.9 1225 1921.5 3014.018 38 62.2 1444 2363.6 3868.84

Sum 185 216.5 5423 6629.9 9015.95 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 8 185 216.5b m+ = 185 5423 6629.9b m+ = Using the methods shown earlier in the exercises or calculator or software regression packages we find the regression equation waste generated to be

1.418 5.727y x= − , where the variable y is amount of waste recovered in millions of tons, and x is the year since 1960. To find the correlation coefficient use the formula, and substitute the values from the table as shown at the top of the next page.







( ) ( ) ( )( ) ( ) ( ) ( )2 2

8 6629.9 185 216.5

8 5423 185 8 9015.95 216.5r

−=

− −

0.854r =

The correlation coefficient between amount of waste recovered and the number of years since 1960 is

0.854r = . b. The solution to this problem is found by setting the equation you found in part (a) for waste recovery equal to .5 times the equation you found in part (b) for waste generation. In other words solve

( )1.418 5.727 .5 3.574 84.626x x− = +

1.418 5.727 1.787 42.313x x− = + .369 48.04x = −

130.2x = − If you could use this model to predict into the past as well as the future, it predicts that the last time we recovered half as much waste as we generated was the year 1830. An intuitive solution to the previous problem would be to recognize that the slope for the equation that models the generation of waste is larger than the slope that models the recovery of waste. This means that instead of recovering a greater percentage of generated waste each year, we are actually recovering a smaller percentage of waste each year. c. Substitute 50x = into each regression equation. The waste generation regression equation predicts that the amount waste generated in 2010 will be

( )3.574 50 84.626 263.326y = + = million tons. The waste recovery regression equation predicts that the amount of waste recovered in 2010 will be

( )1.418 50 5.727 65.173y = − = million tons. 32. a. The scatter plot is plotted at the top of the next column with the regression equation:

x (Max Wingspan in mm)

y (Max length in mm)

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

102030405060708090

100110120130140

It appears that the data is not actually linear. The data appears to curve upwards as x increases. b. Create a chart like the following.

n x y x*x x*y y*y 1 115 80 13225 9200 6400 2 140 90 19600 12600 8100 3 70 75 4900 5250 5625 4 100 75 10000 7500 5625 5 150 130 22500 19500 169006 150 110 22500 16500 121007 135 100 18225 13500 10000

Sum 860 660 110950 84050 64750 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 7 860 660b m+ = 860 110950 84050b m+ = The solution to this system is 0.56m = and 25.479b = The least squares regression line is calculated to be

0.56 25.479y x= + , where y is the maximum caterpillar length, in mm, and x is the maximum wingspan, in mm. c. Using the correlation coefficient formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 2

7 84050 860 660

7 110950 860 7 64750 660r

−=

− −

The correlation coefficient is 0.811r = . This correlation coefficient describes how well the linear regression line fits the data. A correlation coefficient of 0.81 implies that the line is a good fit to describe the relationship between the maximum wingspan and the maximum length of the caterpillar, but normally you would like to see a higher coefficient.






Chapter 1 Summary Exercises 75

d. Substitute 150y = into the regression equation to get 150 0.56 25.479x= + . Solving this equation for x yields:

222.36x = The model predicts the maximum wingspan of a giant silkworm moth that had a maximum caterpillar length of 150 mm would be 222.36 mm. e. Substitute 70x = into the regression equation to get ( )0.56 70 25.479 64.679y = + = The regression model would predict a caterpillar with a maximum wingspan of 70 mm would have a maximum caterpillar length of 64.679 mm. This leads us to believe that the linear model is not a good predictor of the relationship between the maximum caterpillar length and maximum wingspan. Chapter 1 Summary Exercises 1. Substitute 0y = into the equation to find the

x -intercept ( )3 5 0 30x − = . Solving this equation for

x by dividing by 3 yields 10x = . The x -intercept is the point ( )10,0 . Substitute 0x = into the equation to find the y -intercept

( )3 0 5 30y− = . Solving this equation for y by dividing

by -5 yields 6y = − . The y -intercept is the point

( )0, 6− . 2. Marginal cost is the additional cost associated with producing an additional item. For linear functions marginal cost is associated with the slope of the line. The marginal cost for this function is $7. Fixed cost is the cost that does not change regardless of output. For linear functions it is associated with the constant term of the function. The fixed cost for this function is $430. 3. The horizontal line that passes through the point ( )3,8− will have the equation 8y = . The vertical line that passes through the point ( )3,8− will

have the equation 3x = −

4. Solve the equation for y to get the equation in slope-intercept form. 5 2 13x y+ = subtract 5x 2 13 5y x= − divide by 2

13 52 2y x= − .

The coefficient in front of the x variable is the slope. The

slope of this equation is 5

2m −= .

5. Substitute the expression ( )2x + into the equation for x to find the y-coordinate.

3 4y x= + substitute ( )2x x= +

( )3 2 4y x= + + simplify the right hand side

3 6 4y x= + + 3 10y x= + .

6. ( ) ( )2 55 1

3f

−− = +

( ) 105 1

3f

−− = +

( ) 75

3f

−− =

7. 5

22 5 10 5y x y x= + ⇒ = + The slope of the line that

is parallel to 2 5 10y x= + is 52

m = . Using point-slope

formula, the equation of the line that passes through

( )0,0 with slope 52

m = is

( )520 0y x− = −

52y x= .

8. Solve 7x y+ = for x by subtracting y .

7x y= − . Now substitute this into 2 5x y+ = .

( )2 7 5y y− + =

14 2 5y y− + = 14 5y− =

9y− = − 9y =

The solution is continued at the top of the next page.







Now substitute the value of y back into the original equation to get the value of x .

( )9 7x + =

2x = − . The point of intersection is ( )2,7− . 9. Add the two equations together to eliminate the y variable. 3 6 135 6 11

8 0 24

x yx y

x y

+ =+ − =

+ =

Solve this equation for x .

24 38

x = = .

Now substitute the value of x back into one of the original equations to get the value for y .

( )3 3 6 13y+ =

6 4y = 23

y = .

The point of intersection is ( )2

33, . 10. The slopes of the two equations are equal, but the y -intercept of 5 3 9x y+ = is ( )0,3 while the

y -intercept of 10 6 10x y+ = is ( )530, . Therefore the

lines are parallel and never intersect. They are inconsistent and have no solution. 11. The slopes of the two equations are equal. The y -intercept of 5 3 9x y+ = is ( )0,3 while the

y -intercept of 10 6 18x y+ = is ( )0,3 as well. Therefore the lines are equivalent, thus the common points are all the points that satisfy5 3 9x y+ = . 12. Create the table as shown. We have four points.

n x y x*x x*y y*y 1 2 8 4 16 64 2 3 6 9 18 36 3 4 5 16 20 25 4 5 9 25 45 81

Sum 14 28 54 99 206

Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 4 14 28b m+ = 14 54 99b m+ = To solve this system multiply the first equation by 14

4− and

add it to the second equation to get 5 1m = which implies that 0.2m = . Substitute this value back into one of the original equations to get ( )4 14 0.2 28b + = Solve this equation for b . 4 2.8 28b + = subtract 2.8 4 25.2b = divide by 4

6.3b = Therefore the least squares regression line for these data points is 0.2 6.3y x= + Looking at a scatter plot of the data we see that the data is not in a nice line. This leads us to believe that the correlation coefficient would be closer to zero then to one or negative one.

x

y

-1 0 1 2 3 4 5 6 7-1

1

2

3

4

5

6

7

8

9

13. a. Substitute 60x = into the cost function

( ) ( )60 7.9 60 2520 2994C = + = . The total cost of producing 60 items is $2994. b. Take the total cost of producing 60 items in found in part

(a) and divide by 60. 2994 49.960

= The average cost of

producing the first 60 items is $49.90. c. The marginal cost is the additional cost of producing the 61st item. The marginal cost is $7.90. (the slope of the linear cost function)






Chapter 1 Summary Exercises 77

14. a. Let x be the number of years since 1990, and let y bet the sales of the discount store in millions of dollars. The 1.8 million dollars in sales in 1990 can be formulated as the point ( )0,1.8 and the 2.7 million dollars in sales in 1992

can be formulated as the point ( )2,2.7 . Use these two points to calculate the slope of the linear function

2.7 1.8 .452 0

m −= =

−.

Now use the slope and the y -intercept, ( )0,1.8 , to find the equation of the line that will estimate sales in future years is .45 1.8y x= + . Where x and y are defined as above. b. Substitute 20y = into the function found in part (a) to get 20 .45 1.8x= + and solve for x 20 .45 1.8x= + 18.2 .45x=

40.44x = It will take 40.44 years to reach 20 million dollars. Therefore the first year sales can be expected to reach 20 million dollars is the year 2031 (1990+41=2031). 15. a. Profit is revenue minus cost. ( ) ( ) ( )P x R x C x= −

( ) ( )65 25 4000P x x x= − +

( ) 40 4000P x x= − b. The break-even point is when ( ) 0P x = .

40 4000 0x − = 40 4000x =

100x = The company will break-even when they sell 100 items. c. The functions are graphed at the top of the next column.

x (# of items)

$

0 20 40 60 80 100 120 140

-4000

-3000

-2000

-1000

1000

2000

3000

4000

5000

6000

7000

8000

9000

(100,6500)

(100,0)

R(x)

C(x)

P(x)

16. a. Create a table like the one shown below:

n x y x*x x*y y*y 1 2 5 4 10 25 2 4 8 16 32 64 3 5 14 25 70 196 4 6 15 36 90 225 5 3 6 9 18 36 6 4 9 16 36 81 7 7 17 49 119 289 8 9 21 81 189 441 9 5 13 25 65 169 10 12 25 144 300 625 11 15 33 225 495 1089 12 9 19 81 171 361

Sum 81 185 711 1595 3601 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 12 81 185b m+ = 81 711 1595b m+ = To solve this system multiply the first equation by 6.75− and add it to the second equation to get 164.25 346.25m = which implies that 2.108m = . Substitute this value back into one of the original equations to get ( )12 81 2.108 185b + =

Solve this equation for b . 12 170.753 185b + = 12 14.247b =

1.187b = Therefore the least squares regression line for these data points is 2.108 1.187y x= + .







b. To find the correlation coefficient use the formula:

( ) ( ) ( )( ) ( ) ( ) ( )2 22 2

n xy x yr

n x x n y y

−=

− −

∑ ∑ ∑∑ ∑ ∑ ∑


( ) ( ) ( )( ) ( ) ( ) ( )2 2

12 1595 81 185

12 711 81 12 3601 185r

−=

− −

0.987r = . The correlation coefficient is 0.987r = . c. The scatter plot and the regression line are shown below:

x

y

0 2 4 6 8 10 12 14

-4-2

2468

10121416182022242628303234

17. The first thing we notice is the y -intercept is ( )0,4 so that eliminates choices (c), (d), and (e). Now using ( )0,4 and ( )2,8 we see the slope of the relation is

8 4 22 0

m −= =

−. This eliminates answer (a), and

reconfirms the correct choice of answer (b) 2 4y x= + . 18. The correlation coefficient must be a number between

1 1r− ≤ ≤ . This eliminates answers (b) and (c). The line is increasing, which means there is a positive relationship, which eliminates answer (a) and (e). The stronger the positive relationship the closer the correlation is to one. Therefore the answer must be (d) 0.89r = .

Sample Test Answers 1. The points are plotted below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

(2,7)

(3, -2)

(-5,0)

(-4,-1)

(-2,6)

2. The equations are graphed below:

x

y

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-9-8-7-6-5-4-3-2-1

123456789

y=x-2

y=2x-4y=3x-6

3. Use the slope-intercept formula to find the equation. The slope is the per unit change or 3.7m = . The equation of the line will be 3.7 2.8y x= − . 4. Let x be the number of items and let y be the total cost of production. The two points associated with this relation are ( )50,650 and ( )70,714 . The slope of the line is

714 650 3.270 50

m −= =

−. Now use the point slope formula

to find the equation of the line. ( )650 3.2 50y x− = −

650 3.2 160y x− = − 3.2 490y x= +






Chapter 1 Sample Test Answers 79

5. Let x be the number of years from the current year and let y be the population of the town.. If the town is growing at a rate of 250 per year, the slope of the line is 250m = . The current population (population when 0x = ) is 2300, which is the y -intercept. Using the slope-intercept formula, the linear function that will give the approximate population x years from now is

250 2300y x= + . 6. The equation of the vertical line that passes through the point ( )2,7− will be 2x = − . 7. Solve the equation for y .

2 9x y− = 2 9y x− = −

92xy −

=−

912 2y x= − .

The equation is now in slope-intercept form. The slope is

12m = and the y -intercept is ( )9

20, − .

To find the x -intercept set 0y = in the original equation.

( )2 0 9x − =

9x = . The x -intercept is ( )0,9 8. The two points on the line are ( )2,0− and ( )0,6.4 .

The slope of the line is 0 6.4 3.2

2 0m −= =− −

. Use the slope

and the y -intercept in the slope-intercept formula to get the equation 3.2 6.4y x= + . 9. Let x be the price of the special, and y be the number of

specials sold. Use the points ( )4.75,78 and ( )5.50,63 .

The slope of the line is 63 78 20

5.50 4.75m −= = −

−.

Use the point-slope formula to find the equation of the line. The calculations are shown below:

( )78 20 4.75y x− = − −

78 20 95y x− = − + 20 173y x= − + .

The equation that will model the number of lunch specials y sold for a given price x is 20 173y x= − + .

10. Let x be the number of years after the purchase of the tractor, and let y be the value of the tractor. Two points

associated with this line are ( )0,5600 and ( )10,2100 .

The slope of the line is 5600 2100 350

0 10m −= = −

−.

Use the slope and the y -intercept to find the equation of the line that models the value of the tractor as a function of the number of years from the date of purchase. The function is 350 5600y x= − + . 11. The slopes of the two equations are different; therefore there is only one intersection point. 12. Substitution method Solve 5x y+ = for x .

5x y= − Substitute x into 2 12x y− =

( )2 5 12y y− − =

10 2 12y y− − = 3 2y− =

23y −=

Now substitute the value of y back into one of the original equations to get x .

( )23 5x −+ =

173x =

The solution is ( )17 23 3, −

Elimination method Add the two equations together to eliminate y

52 12x yx y+ =

+ − =

3 0 17x y+ = Solve for x 3 17x =

173x =








Substitute x back into one of the original equations to find y .

173 5y+ =

23y −= .

The solution is ( )17 23 3, −

13. Solve both equations for y . Equation 1 0.2 0.8 3x y+ = 0.8 0.2 3y x= − +

1514 4y x−= +

Equation 2

4 15x y+ = 4 15y x= − +

1514 4y x−= +

The slope of both equations is 1

4m −= . The y-intercept of

both equations is ( )1540, . Therefore there are infinitely

many solutions. All points given by the equation 151

4 4y x−= + will solve the system. 14. Revenue is price times quantity. The revenue function is

( ) 98R x x= . The cost function is direct cost plus fixed cost. The Cost function is ( ) 73 40,000C x x= + . Profit is revenue minus cost. The profit function will be ( ) ( ) ( )P x R x C x= −

( ) ( )98 73 40,000P x x x= − +

( ) 25 40,000P x x= − To find the break-even point set ( ) 0P x = and solve for x .

25 40,000 0x − = 25 40,000x =

1600x = The company will break-even if they sell 1600 calculators. 15. To find the equilibrium price and quantity set supply equal to demand and solve for p as shown at the top of the next column.

( ) ( )S p D p=

0.03 9 17.5 0.015p p− = − Solve for p . 0.045 9 17.5p − = 0.045 26.5p =

588.89p = Substitute the price back into supply or demand to get the equilibrium quantity. ( ) ( )588.89 0.03 588.89 9 8.67S = − = .

Equilibrium is reached when the price is $588.89. At this price the annual supply is 8.67 quilts, and the annual demand for quilts is 8.67. 16. Let x start with year 1, be the year the mutual fund was invested, and let y be the return (in percentage terms)

achieved by the mutual fund ( )4 4%y return= ⇒ . a. Using the above variables four data points will be ( )1,8 ; ( )2,9 ; ( )3,12 ; ( )4,14 over the first four years of the fund. These points are plotted on the axis below:

x (# of years)

y (%return on investment)

0 1 2 3 4 5 6 7 8 9-1

123456789

101112131415161718192021222324

b. Create a chart like the following using the data points.

n x y x*x x*y y*y 1 1 8 1 8 64 2 2 9 4 18 81 3 3 12 9 36 144 4 4 14 16 56 196

Sum 10 43 30 118 485 Let m and b represent the slope and y-intercept of the least squares regression line. The system of equations that give us these values is 4 10 43b m+ = 10 30 118b m+ =






Chapter 1 In–Depth Application 81

To solve the system on the previous page multiply the first equation by 10

4− and add it to the second equation to get

5 10.5m = which implies that 2.1m = . Substitute this value back into one of the original equations to get ( )4 10 2.1 43b + =

Solve this equation for b . 4 21 43b + = subtract 21 4 22b = divide by 4

5.5b = Therefore the least squares regression line for these data points is 2.1 5.5y x= + . c. To find the correlation coefficient use the formula from the text and substitute the appropriate values from the table.

( ) ( ) ( )( ) ( ) ( ) ( )2 2

4 118 10 43

4 30 10 4 485 43r

−=

− −

0.984r = The correlation coefficient is 0.984r = . d. Substitute 5x = into the regression equation to get

( )2.1 5 5.5 16y = + = . The regression model predicts a 16% return on investments for the fifth year. e. Substitute 20y = into the regression equation and solve for x .

( )2.1 5.5 20x + =

2.1 14.5x = 6.9x =

The regression model predicts that in the 7th year the fund will return 20% on the investment.

In–Depth Application: Unemployment and Homicides in the

United States 1. Let x be the number of years since 1980 and let y be the number of unemployed per 100 people. The scatter plot for the data is shown at the top of the next column.


y (Unemployed per 100)

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

123456789

1011121314

From the plot we can see unemployment rose from 1980 to 1982 ( 0 2x< < ) and in the years 1989 to 1992 ( 9 12x< < ). Unemployment was falling in the years 1982 to 1989 ( 2 9x< < ) and again in the years 1992 to 1999 (12 19x< < ) 2. Let x be the number of years since 1980 and let y be the number of homicides per 1000,000 people. The scatter plot for the data is shown below:


y (Homicides per 100000)

-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

123456789

1011121314

From the plot we can see that the homicide rate generally increased in the years 1984 to 1991( 4 11x< < ). In 1987 there was a one year decline, but the rates continued to increase after that. Homicide rates were falling in the years 1980 to 1984 ( 0 4x< < ) and in the years 1991 to 1999 (11 19x< < ) In 1993 we see a one year increase, but the rates continued to decrease after that.







c. It appears that unemployment rates and homicide rates have the same pattern. When unemployment rates are falling, so are the homicide rates. This is not an exact match, but it looks close. d. First let x be the unemployment rate, and y be the number of homicides per 100,000. Create a chart like before and use the correlation coefficient formula to find

n x y x*x x*y y*y 1 7.1 10.22 50.41 72.562 104.44842 7.6 9.83 57.76 74.708 96.62893 9.7 9.07 94.09 87.979 82.26494 9.6 8.25 92.16 79.2 68.06255 7.5 7.91 56.25 59.325 62.56816 7.2 7.95 51.84 57.24 63.20257 7 8.55 49 59.85 73.10258 6.2 8.26 38.44 51.212 68.22769 5.5 8.41 30.25 46.255 70.7281

10 5.3 8.66 28.09 45.898 74.995611 5.6 9.42 31.36 52.752 88.736412 6.8 9.79 46.24 66.572 95.844113 7.5 9.31 56.25 69.825 86.676114 6.9 9.51 47.61 65.619 90.440115 6.1 8.96 37.21 54.656 80.281616 5.6 8.22 31.36 46.032 67.568417 5.4 7.41 29.16 40.014 54.908118 4.9 6.8 24.01 33.32 46.24 19 4.5 6.28 20.25 28.26 39.438420 4.2 5.7 17.64 23.94 32.49

Sum 130.2 168.51 889.38 1115.219 1446.852 The correlation coefficient calculated from the formula is 0.542r = . This is a positive correlation, but it is not very high. There does not appear to be a significant relationship between unemployment rates and homicide rates in the US over this time period. 5. For the years 1980 to 1988, create a chart using that data.

n x y x*x x*y y*y 1 7.1 10.22 50.41 72.562 104.4484 2 7.6 9.83 57.76 74.708 96.6289 3 9.7 9.07 94.09 87.979 82.2649 4 9.6 8.25 92.16 79.2 68.0625 5 7.5 7.91 56.25 59.325 62.5681 6 7.2 7.95 51.84 57.24 63.2025 7 7 8.55 49 59.85 73.1025 8 6.2 8.26 38.44 51.212 68.2276 9 5.5 8.41 30.25 46.255 70.7281

Sum 67.4 78.45 520.2 588.331 689.2335

The correlation coefficient is 0.09r = which implies that there is almost no linear relationship between the unemployment rate and the homicide rate. 6. For the years 1989 to 1999 create a chart like the following.

n x y x*x x*y y*y 1 5.3 8.66 28.09 45.898 74.99562 5.6 9.42 31.36 52.752 88.73643 6.8 9.79 46.24 66.572 95.84414 7.5 9.31 56.25 69.825 86.67615 6.9 9.51 47.61 65.619 90.44016 6.1 8.96 37.21 54.656 80.28167 5.6 8.22 31.36 46.032 67.56848 5.4 7.41 29.16 40.014 54.90819 4.9 6.8 24.01 33.32 46.24 10 4.5 6.28 20.25 28.26 39.438411 4.2 5.7 17.64 23.94 32.49

Sum 62.8 90.06 369.18 526.888 757.6188 The correlation coefficient calculated from the formula is

0.87r = this is a very strong positive correlation. The data suggest that there is a strong relation between the unemployment and the homicide rate over the years 1989 to 1999. 7. There are an unlimited number of factors that could influence the trends in the data. Economic reasons could vary from the recession in the mid 80’s to the technology bubble of the 90’s. Political events have seen Republicans occupy the White House from 1980–1992, and Democrats occupy the White House from 1992–1999. There have been several conflicts including the Gulf War. All of these factors and many more will influence the data that we have.






Documents

Solution Manual for Finite Mathematics An Applied Approach ... · Solution Manual for Finite Mathematics An Applied Approach 3rd Edition by Young Full file at ... Solution Manual