hw6 - TARIQ ELHASSANI Homework #6 Problem 13.5 Linregr (x,...

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TARIQ ELHASSANI Homework #6 Problem 13.5 Linregr (x, y) a = 0.3525 G the slope St = 55.6000 Sr = 9.0740 r = 0.9148 Intercept = 4.8515 Graph 0 2 4 6 8 10 12 14 16 18 20 4 5 6 7 8 9 10 11 12 B) switching the variables we get Linregr (y, x) a = 2.3741 + the slope St = 374.5000 Sr = 61.1187 r = 0.9148 Sxy= 1.0650 Intercept = -9.9676
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5 6 7 8 9 10 11 12 0 2 4 6 8 10 12 14 16 18 20 conclusion Problem 13.11 Linregr (log10 (w), log10 (A)) r2 = 0.9711 Ans = 0.3799 -0.3821 The slope is 0.3799 The intercept is –0.3821 Therefore • log (A) = log (a) + b* log (w) ± b= 0.3799 Calculation of the area a = 0.4149 b = 0.3799 W= 95 Kg A= a*W^b A= 2.34 m^2
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Problem 14.2 The first part, see the derivations in the attached papers prob142(x,y) sx2 = 20400 sxy = 312850 sy2 = 5104325 sx3 = 1296000 sx4 = 87720000 sx2y = 20516500 Sr = 2.3016e+005 Ans = 7.7710 0.1191 0 = The best fit is: y=7.7710+0.1191x Graph 0 10 20 30 40 50 60 70 80 90 100 0 200 400 600 800 1000 1200 1400 1600 1800 2000
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This note was uploaded on 10/17/2010 for the course ME 335 taught by Professor Chaturvedi during the Spring '10 term at Old Dominion.

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hw6 - TARIQ ELHASSANI Homework #6 Problem 13.5 Linregr (x,...

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