careyexcelexercises

# careyexcelexercises - Is it not really worthwhile plotting...

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X Y 2nd Y Percent Difference 2.5 2.72 2.7 0.58 5.1 1.76 1.81 -2.53 7.6 1.3 1.27 2.68 8.9 1.13 1.11 1.68 10.2 1.02 1.04 -2.16 The greatest percent difference observed is 2.68% 0 2 4 6 8 10 12 0 2 4 6 8 10 12

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Yes, X=7.0 is within the range of the approximation data. From the graph, the estimate of x=7 looks to be about 1.4. The actual value of x=7 from the 2nd order polynomial expression is 1.345.  Yes, the value of Y appears to be close to the "exact" value given in the table. Exercise 2 X 1/Y 2nd Y Percent Difference 2.5 0.37 0.36 0.95 5.1 0.57 0.57 -0.86 7.6 0.77 0.77 -0.52 8.9 0.89 0.88 0.85 10.2 0.98 0.98 -0.17 0 2 4 6 8 10 12 0 2 4 6 8 10 12
Yes, a straight line relationship seems to be implied. The best fit straight-line (linear) equation for the data, when calculated by the graph trend line function on excel, is y=0.080151x+0.164163 The percent difference from the table value for Y is not that different to the percent difference when using the polynomial fit of exercise 1.

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Unformatted text preview: Is it not really worthwhile plotting data in various ways first before deciding what kind of function will be used to represent the data. 2 4 6 8 10 12 2 4 6 8 10 12 f(x) = 1.#NANx 1/Y Linear Re-gression for Exercise 4 y alpha 0 0.785; -0.785 1 .524; -1.571 2 .375; N/A 3 .285; N/A 2.6 N/A Another Way to Do Exercise 4 y negativesqaure root postive square root positive square root negative square root-0.79 0.79 0.71-0.71 1-1.57 0.52-0.25-1.75 2 #NUM! 0.37-1.13-2.87 3 #NUM! 0.28-1.97-4.03 2.6 #NUM! 0.32-1.64-3.56 0.5 1 1.5 2 2.5 3 3.5-2-1.5-1-0.5 0.5 1 f(x) = -0.16x + 0.74 f(x) = -0.79x - 0.79 negativesqaure root Linear Re-gression for negat-postive square root Linear Regres-sion for postive square root ....
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careyexcelexercises - Is it not really worthwhile plotting...

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