bab 5 regrasi linier dengan metode kuadrat terkecil

n x y x.y1 2 0.35 4 0.72 4.6 0.75 21.16 3.453 7 1.2 49 8.44 9.4 1.55 88.36 14.575 11.5 1.85 132.25 21.275

∑ 34.5 5.7 294.77 48.395

a= 0.1642

y= 0.1642x

latihan 2berikut ini adalah data-data konsentrasi vs waktu untuk reaksi orde 1

ca (gmol/liter) 0.8 0.75 0.6723 0.512 0.4627 0.3086t(detik) 0 100 200 200 400 500

nilai k konstanta kecepatan reaksi

n x y x.y1 0 0 0 0 a= -0.0017 = -k2 100 -0.064539 10000 -6.4538523 200 -0.173907 40000 -34.781414 200 -0.446287 40000 -89.257425 400 -0.547533 160000 -219.01316 500 -0.952566 250000 -476.2829

∑ 1400 -2.184831 500000 -825.7887jadi nilai k adalah 0.0017

n x y x.y1.0000 2.5000 2.5000 6.2500 6.25002.0000 3.2000 4.6000 10.2400 14.72003.0000 1.5000 5.7500 2.2500 8.62504.0000 2.7500 7.0200 7.5625 19.30505.0000 3.3500 8.3300 11.2225 27.9055

x2

y=a0.x

x2

x2

xay

tkc

c

A

A

.0

0

.ln

6.0000 4.5000 9.7000 20.2500 43.65007.0000 3.5500 10.1000 12.6025 35.8550

∑ 21.3500 48.0000 70.3775 156.3105

7 a0+ a1* 21.3500 = 48.0000 * 21.350021.3500 a0+ a1* 70.3775 = 156.3105 * 7

7 a0+ 40.22608 = 48.0000a0 = 1.1106

y=a0+a1xy= 1.1106 + 1.8841 x

n x y x.y1.0000 2.8000 1.1400 7.8400 3.19202.0000 4.0000 1.5400 16.0000 6.16003.0000 5.1000 2.5600 26.0100 13.05604.0000 7.5000 3.4000 56.2500 25.50005.0000 8.8000 4.3000 77.4400 37.84006.0000 10.6000 5.9000 112.3600 62.54007.0000 14.0000 7.1000 196.0000 99.4000

∑ 52.8000 25.9400 491.9000 247.6880

7 a0+ a1* 52.8000 = 25.9400 * 52.800052.8000 a0+ a1* 491.9000 = 247.6880 * 7

7 a0+ 29.33652 = 25.9400a0 = -0.4852

y=a0+a1xy= -0.4852 + 0.5556 x

x2

149.45 a0+ a1* 455.8225 = 1024.8000149.45 a0+ a1* 492.6425 = 1094.1735

-a1* -36.82 = -69.3735

a1 = 1.8841

369.6 a0+ a1* 2787.84 = 1369.6320369.6 a0+ a1* 3443.3 = 1733.8160

-a1* -655.46 = -364.1840

a1 = 0.5556