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The table shows a meteorologist's predicted temperatures for an April day in Washington D.C. Use quadratic regression to find a quadratic model for this data. (Use the 24-hour clock to represent times after noon.)


Time

Predicted Temperature


8 a.m.

50.68


10 a.m.

62.24


12 p.m..

69.75


2 p.m..

73.2


4 p.m...

72.59


6 p.m..

67.93


a.) –0.007x2 + 15.907x – 40.131

b.) –0.507x2 + 14.907x – 36.131

c.) –0.507x2 – 14.907x + 36.131

d.) –1.007x2 + 13.907x – 32.131

Verified Answer :

nobillionaireNobley

–0.507x2 + 14.907x – 36.131

The correct answer is:


b.) –0.507x2 + 14.907x – 36.131


Explanation:


First we enter the data into a graphing calculator. The times are the x-coordinates and the temperatures are the y-coordinates; this is because temperature is dependent upon time.


After we enter this, we run a regression on the calculator. We will run the quadratic regression; the equation is of the form

y = ax²+bx+c.


The values we are given on a Casio fx-9860 gii are:

a = -0.507

b = 14.908

c = -36.136


This is closest to choice B.

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