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question 16 of 25 what is the quadratic regression equation that fits t…

Question

question 16 of 25
what is the quadratic regression equation that fits these data?
number of seconds | height (in feet)
0 | 12
1 | 13
2 | 15
3 | 13
4 | 9
5 | 7
a. $y=14.77(0.89^{x})$
b. $y=-2.04x^{2}+1.46x+13.92$
c. $y=-0.70x^{2}+2.37x+11.96$
d. $y=-1.04x^{2}+3.04x+9.68$

Explanation:

Step1: Test x=0 in each option

For A: $y=14.77(0.89^0)=14.77$, does not match 12.
For B: $y=-2.04(0)^2+1.46(0)+13.92=13.92$, does not match 12.
For C: $y=-0.70(0)^2+2.37(0)+11.96=11.96\approx12$, matches.
For D: $y=-1.04(0)^2+3.04(0)+9.68=9.68$, does not match 12.

Step2: Verify x=1 for option C

$y=-0.70(1)^2+2.37(1)+11.96=-0.70+2.37+11.96=13.63\approx13$, matches.

Step3: Verify x=2 for option C

$y=-0.70(2)^2+2.37(2)+11.96=-2.8+4.74+11.96=13.9\approx15$ (close, and closer than others).

Step4: Verify x=3 for option C

$y=-0.70(3)^2+2.37(3)+11.96=-6.3+7.11+11.96=12.77\approx13$, matches.

Step5: Verify x=4 for option C

$y=-0.70(4)^2+2.37(4)+11.96=-11.2+9.48+11.96=10.24\approx9$, close match.

Step6: Verify x=5 for option C

$y=-0.70(5)^2+2.37(5)+11.96=-17.5+11.85+11.96=6.31\approx7$, close match.

Answer:

C. $y=-0.70x^{2}+2.37x+11.96$