Approximation and Errors33.1Significant Figures3.2Scientific Notation3.3Approximation and ErrorsCase StudyChapter Summary

Suppose that there are actually 116 paper clips on the plate.Case StudyJohns estimate Actual number 120 116Did John or Linda make a better estimation?We need to find the difference between each estimate and the actual number.The smaller the difference is, the better is the estimate. 4Actual number Lindas estimate 116 110 6Since 4 is less than 6, John made a better estimation.

3.1 Significant FiguresA. Basic ConceptsThe population of Hong Kong at the end of 2007 is 7 000 000. (correct to the nearest million) 6 950 000. (correct to the nearest ten thousand) 6 953 000. (correct to the nearest thousand)In most cases, the far left non-zero digit, having the largest place value in a number, is the first significant figure.Subsequent important digits are called the second significant figure, the third significant figure, and so on.The important digits of a number are called significant figures.

7 6 95 6 953This is also the most important figure.

For all numbers, any zeros between 2 non-zero digits are significant figures.For example, the number 1049 has 4 significant figures.3.1 Significant FiguresA. Basic Concepts

3.1 Significant FiguresA. Basic ConceptsFor all decimals, any zeros after the last non-zero digit are significant figures.For decimals less than 1, all zeros in front of the first non-zero digit are not significant figures. These zeros are called place holders.For example, the number 0.0380 has only 3 significant figures.

The following table shows some examples. All the significant figures are marked with green colour.3.1 Significant FiguresA. Basic Concepts

Using the technique of rounding off for estimation,the number 47 054 can be expressed as:(a)47 100 (cor. to the nearest hundred)(b)47 000 (cor. to the nearest thousand)(c)50 000 (cor. to the nearest ten thousand)Alternatively, we can round off a number to a certain number of significant figures.3.1 Significant FiguresB. Round off to the Required Significant FiguresWe can express 47 054 as:(a)47 100 (cor. to 3 sig. fig.)(b)47 000 (cor. to 2 sig. fig.)(c)50 000 (cor. to 1 sig. fig.)

3.1 Significant FiguresB. Round off to the Required Significant FiguresThe following steps show how to round off 51 672, correct to 3 significant figures.

Round off 472 780 correct to(a)2 significant figures,(b)3 significant figures,(c)4 significant figures.(a)470 000 (cor. to 2 sig. fig.)(b)473 000 (cor. to 3 sig. fig.)(c)472 800 (cor. to 4 sig. fig.)3.1 Significant FiguresB. Round off to the Required Significant FiguresExample 3.1TSolution:

3.1 Significant FiguresB. Round off to the Required Significant FiguresExample 3.2TRound off 0.300 649 correct to(a)1 significant figure,(b)2 significant figures,(c)5 significant figures.Solution:(a)0.3 (cor. to 1 sig. fig.)(b)0.30 (cor. to 2 sig. fig.)(c)0.300 65 (cor. to 5 sig. fig.)

3.1 Significant FiguresB. Round off to the Required Significant FiguresExample 3.3TRound off 9995 correct to(a)1 significant figure,(b)2 significant figures,(c)3 significant figures.Solution:(a)10 000 (cor. to 1 sig. fig.)(b)10 000 (cor. to 2 sig. fig.)(c)10 000 (cor. to 3 sig. fig.)

The total price of 120 oranges is $250. Round off the average price of each orange correct to 2 significant figures.3.1 Significant FiguresB. Round off to the Required Significant FiguresExample 3.4T The average price of each orange $(250 120) $2.0833Solution:

3.2 Scientific NotationA. Introduction Our body produces about 173 000 000 000 red blood cells each day. The mass of 1 water molecule is about 0.000 000 000 000 000 000 000 03 g.It is convenient to rewrite these numbers as follows: 173 000 000 000 1.73 100 000 000 000 1.73 1011This kind of representation of numbers is called scientific notation.In science, we often deal with numbers which are very large or small: 0.000 000 000 000 000 000 000 03 g 3 1023 gA positive number is expressed in scientific notation if it is in the form a 10n, where 1 a 10 and n is an integer.

Express each of the following numbers in scientific notation.(a)22 000(b)0.000 000 7(c)95 000 104(d)0.008 1023.2 Scientific NotationA. IntroductionExample 3.5TSolution:(a)22 000 2.2 10 000(b)0.000 000 7 7 0.000 000 1(c) 95 000 104 (9.5 104) 104(d)0.008 102 (8 103) 102

3.2 Scientific NotationA. IntroductionExample 3.6TConvert the following numbers into integers or decimals.(a)1.002 107(b)6 105 Solution:(a) 1.002 107 1.002 10 000 000(b) 6 105 6 0.000 01

3.2 Scientific NotationA. IntroductionExample 3.7TSolution:

3.2 Scientific NotationA. IntroductionThe calculator would display 1.32E9 before execution.

B. Applications of Scientific NotationFor example:1 228 550 000 m3 of water was consumed in Hong Kong in 2005 2006. 1 228 550 000 m3 1 230 000 000 m3 (cor. to 3 sig. fig.) 1.23 109 m33.2 Scientific NotationWhen presenting estimates, we can express numbers in scientific notation and round off the value correct to a certain number of significant figures.

Example 3.8TB. Applications of Scientific Notation3.2 Scientific Notation36 084 people were injured in road accidents last year. Round off the number correct to 3 significant figures and express the answer in scientific notation.Solution:36 084 36 100 (cor. to 3 sig. fig.)

Example 3.9TB. Applications of Scientific Notation3.2 Scientific NotationThe atomic radii of a helium atom and a gold atom are 3.1 1011 m and 1.35 1010 m respectively. How many times is the atomic radius of a gold atom to that of a helium atom? Give the answer correct to 2 significant figures.Solution:The required ratio(cor. to 2 sig. fig.)

3.3 Approximation and ErrorsThey are only approximations.Hence, every measurement or estimation has errors.In Book 1A Chapter 2, we learnt that no measurements give exact values.

A. Absolute Error3.3 Approximation and ErrorsThe difference between a measured value (or an estimated value) and the actual value is called the absolute error.If the actual value is larger than a measured value, thenabsolute error actual value measured valueIf a measured value is larger than the actual value, thenabsolute error measured value actual value Example:A story book has 117 pages. Sara guessed that the book has 100 pages. The absolute error of Saras estimation is 17 pages.

Example 3.10TA. Absolute Error3.3 Approximation and ErrorsThe actual weight of a pack of potato chips is 183.4 g. Find the absolute error if Julie corrects the weight to 1 significant figure.Solution:183.4 g 200 g (cor. to 1 sig. fig.)The absolute error (200 183.4) g

Therefore, errors of measurements are related to the scale intervals of the tools.In the figure, the scale interval of the ruler is 1 cm.B. Maximum Absolute Error3.3 Approximation and ErrorsMeasuring tools with smaller scale intervals can give estimations with higher accuracy.The minimum and the maximum actual lengths are 14.5 cm and 15.5 cm respectively.The measured length is 15 cm, correct to the nearest cm.The absolute error of this measurement does not exceed 0.5 cm. This figure is called the maximum absolute error.

Lower limit ( 14.5 cm)Minimum possible value of measurement Upper limit ( 15.5 cm)Maximum possible value of measurementThey can be found by the following formulas.B. Maximum Absolute Error3.3 Approximation and ErrorsIn general,In the figure, the scale interval of the ruler is 1 cm.Lower limit Measured value Maximum absolute errorUpper limit Measured value Maximum absolute error

Example 3.11TB. Maximum Absolute Error3.3 Approximation and ErrorsThe capacity of a bottle is 5.38 L, correct to 3 significant figures. What is the lower limit of the capacity?Solution:The lower limit of the capacity (5.38 0.005) L 0.005 LThe scale interval of measurement 0.01 L

Lower limit (8 0.5) cm 7.5 cmUpper limit (8 0.5) cm 8.5 cmLower limit 10.5 cmUpper limit 11.5 cmLower limit of the areaUpper limit of the areaThe actual area lies between 39.375 cm2 and 48.875 cm2.Example 3.12TB. Maximum Absolute Error3.3 Approximation and ErrorsGordon measured the base length and height of a triangle to be 8 cm and 11 cm respectively, both correct to the nearest 1 cm. What is the range of the area of the triangle?Solution:Base length:Height:

Both of the absolute errors of the 2 measurements are the same, but they do not reflect the degree of accuracy.To compare the accuracy of 2 estimations, we also have to determine the relative error.C. Relative Error3.3 Approximation and ErrorsConsider the following set of data:

AnimalActual weight (kg)Measured weight (kg)Absolute error (kg)Pig63.42630.42Dog5.194.770.42

The absolute error of the pigs weight is small (insignificant) when compared to its actual weight.On the other hand, the absolute error of the dogs weight is significant because it is relatively large when compared to its actual weight.In general, we use the relative error to determine the degree of accuracy of a measurement.C. Relative Error3.3 Approximation and Errors

For example, the degree of accuracy of the measurement of the pigs weight is higher than that of the dogs weight.Sometimes it is impossible to find the actual values in some real-life situations.So we can use the maximum absolute error and the measured value instead to find the relative error.C. Relative Error3.3 Approximation and ErrorsThe smaller the relative error is, the higher is the accuracy of a measurement.

The figure below shows the length of a piece of string. Find(a)the length of the string,(b)the maximum absolute error of the length,the relative error of the length, correct to 3 significant figures.(b)Maximum absolute error(c)Relative error(cor. to 3 sig. fig.) Example 3.13TC. Relative Error3.3 Approximation and ErrorsSolution:

Let x g be the weight of a package of sugar.x 2.5 72 180The weight of a package of sugar is 180 g.Example 3.14TC. Relative Error3.3 Approximation and ErrorsSolution:

For example, since the relative errors of the pigs and the dogs weight are 0.006 62 and 0.0809 respectively.The percentage errors of the pigs and the dogs weight are 0.662% and 8.09% respectively.The smaller the percentage error is, the higher is the accuracy of a measurement.D. Percentage Error3.3 Approximation and ErrorsWhen the relative error is expressed as a percentage, it is called the percentage error.Percentage error Relative error 100%

A university bought 1258 new computers last year. If the number of computers bought is now correct to the nearest hundred, find thepercentage error of the estimation. (Give the answer correct to 3 significant figures.)Absolute error 1300 1258 42Percentage error (cor. to 3 sig. fig.)Example 3.15TD. Percentage Error3.3 Approximation and ErrorsSolution:1258 1300 (cor. to the nearest hundred)

Sally and Christine measured the capacity of 2 boxes separately. Sallys result was 300 mL correct to 2 significant figures. Christines result was 1250 mL correct to the nearest 50 mL.(a)Find the percentage errors of their measurements. (Give the answer correct to 3 significant figures if necessary.)(b)Hence determine who measured more accurately.(cor. to 3 sig. fig.) (a)For Sallys measurement: Example 3.16TD. Percentage Error3.3 Approximation and ErrorsSolution:Maximum absolute error10 mL 2 5 mLPercentage errorFor Christines measurement: Maximum absolute error50 mL 2 25 mLPercentage error(b)Since 1.67% is less than 2%, Sally measured more accurately.

Chapter Summary3.1 Significant FiguresFor all numbers, any zeros between 2 non-zero digits are significant figures.For all decimals, any zeros after the last non-zero digit are significant figures.

3.2 Scientific NotationAll positive numbers can be expressed in the form a 10n, where 1 a 10 and n is an integer.Chapter Summary

3.3 Approximation and Errors1. Absolute error (a)If the actual value is larger than a measured value, then the absolute error actual value measured value. (b)If a measured value is larger than the actual value, then the absolute error measured value actual value.Chapter Summary(b)Lower limit Measured value Maximum absolute error(c)Upper limit Measured value Maximum absolute error

3.3 Approximation and ErrorsChapter SummaryPercentage errorPercentage error Relative error 100%3. Relative error

B. Round off to the Required Significant Figures3.1 Significant FiguresFollow-up 3.1Round off 94 506 correct to(a)1 significant figure,(b)2 significant figures,(c)3 significant figures.(a)90 000 (cor. to 1 sig. fig.)(b)95 000 (cor. to 2 sig. fig.)(c)94 500 (cor. to 3 sig. fig.)Solution:

B. Round off to the Required Significant Figures3.1 Significant FiguresFollow-up 3.2Round off 0.203 617 correct to(a)2 significant figures,(b)3 significant figures,(c)4 significant figures.(a)0.20 (cor. to 2 sig. fig.)(b)0.204 (cor. to 3 sig. fig.)(c)0.2036 (cor. to 4 sig. fig.)Solution:

B. Round off to the Required Significant Figures3.1 Significant FiguresFollow-up 3.3Round off 89 989 correct to(a)1 significant figure,(b)2 significant figures,(c)4 significant figures.(a)90 000 (cor. to 1 sig. fig.)(b)90 000 (cor. to 2 sig. fig.)(c)89 990 (cor. to 4 sig. fig.)Solution:

Total weight of the mobile phones (85.53 3) g 256.59 gB. Round off to the Required Significant Figures3.1 Significant FiguresFollow-up 3.4The weight of each of 3 mobile phones is 85.53 g. Round off the total weight of the mobile phones correct to 3 significant figures.Solution:

Follow-up 3.53.2 Scientific NotationA. IntroductionExpress each of the following numbers in scientific notation.(a)1 475 000(b)0.000 000 061(c)28 600 000(d)1300 104Solution:(a)1 475 000 1.475 1 000 000(b)0.000 000 061 6.1 0.000 000 01(c) 28 600 000 2.86 10 000 000(d)1300 104 (1.3 103) 104

Follow-up 3.63.2 Scientific NotationA. IntroductionConvert the following numbers into integers or decimals.(a)4.6 109(b)5.02 104Solution:(a) 4.6 109 4.6 1 000 000 000(b) 5.02 104 5.02 0.0001

Follow-up 3.73.2 Scientific NotationA. IntroductionSolution:the same as 2 104

Follow-up 3.8B. Applications of Scientific Notation3.2 Scientific NotationA flower shop made $764 280 in profit last year. Round off thenumber correct to 3 significant figures and express the answer inscientific notation.Solution:$764 280 $764 000 (cor. to 3 sig. fig.)

Follow-up 3.9B. Applications of Scientific Notation3.2 Scientific NotationThe mass of the Moon is about 7.35 1022 kg. It is 1.23 102 times as heavy as the Earth. Find the mass of the Earth. Express the answer in scientific notation and correct to 3 significant figures.Solution:The mass of the Earth(cor. to 3 sig. fig.)

Follow-up 3.10A. Absolute Error3.3 Approximation and ErrorsThe actual capacity of a cup is 751.5 mL. Find the absolute error if Jane corrects the value of the capacity to 3 significant figures.Solution:751.5 mL 752 mL (cor. to 3 sig. fig.)The absolute error (752 751.5) mL

Follow-up 3.11B. Maximum Absolute Error3.3 Approximation and ErrorsThe weight of a ring is measured to be 45.7 g, correct to 3 significant figures. What is the lower limit of the weight?Solution:The lower limit of the weight (45.7 0.05) g 0.05 gThe scale interval of measurement 0.1 g

Follow-up 3.12B. Maximum Absolute Error3.3 Approximation and ErrorsTiffany measured the length of the sides of an isosceles triangle. The results are shown in the figure. If the measurements were correct to the nearest 2 cm, find the range of the perimeter of the triangle.Solution:Lower limit (20 1) cm 19 cmUpper limit (20 1) cm 21 cmLower limit 23 cmUpper limit 25 cmLower limit of the perimeter (19 + 23 + 23) cm 65 cmUpper limit of the perimeter (21 + 25 + 25) cm 71 cmThe range of the perimeter is 65 cm 71 cm.Base length:Length of the sides 24 cm:

Follow-up 3.13C. Relative Error3.3 Approximation and ErrorsThe figure shows the thermometer in a classroom. Find(a)the temperature in the classroom,(b)the maximum absolute error of the temperature,(c)the relative error of the temperature, correct to 3significant figures.Solution:(cor. to 3 sig. fig.)

Follow-up 3.14C. Relative Error3.3 Approximation and ErrorsLet x s be the actual result. 62.5The actual result is 62.5 s.Jimmy recorded the time for him to finish a 400 m run with a maximum absolute error of 0.01 s. If the relative error of the result is 1.6 104, find the lower limit of the actual result.Solution:

Follow-up 3.15D. Percentage Error3.3 Approximation and Errors364 students joined a charity event. If the number of students is now correct to the nearest ten, find the percentage error of the estimation. (Give the answer correct to 2 significant figures.)Solution:Absolute error 364 360 4Percentage error (cor. to 2 sig. fig.)364 students 360 students (cor. to the nearest ten)

Follow-up 3.16D. Percentage Error3.3 Approximation and ErrorsKelvin and Florence measured the durations of 2 fire drills separately. Kelvins result was 210 seconds correct to the nearest 10 seconds. Florences result was 360 seconds correct to the nearest 20 seconds.(a)Find the percentage errors of their measurements.(Give the answers correct to 3 significant figures.)(b)Hence determine who measured more accurately.Solution:(cor. to 3 sig. fig.) (a)For Kelvins measurement: Maximum absolute error10 s 2 5 sPercentage errorFor Florences measurement: Maximum absolute error20 s 2 10 sPercentage error(b)Since 2.38% is less than 2.78%, Kelvin measured more accurately. (cor. to 3 sig. fig.)