Question

question 21 of 25
what is the exponential regression equation that fits these data?
a. $y=104.74x - 206.6$
b. $y=1.31(2.80^{x})$
c. $y=39.16x^{2}-169.38x + 158.9$
d. $y=2.80(1.31^{x})$

Explanation:

Step1: Recall exponential regression form

Exponential regression has the form $y = ab^x$, where $a>0, b>0, b
eq1$. Eliminate non-exponential options (A is linear, C is quadratic).

Step2: Test option B with $x=1$

Substitute $x=1$ into $y=1.31(2.80^x)$:
$y=1.31\times2.80^1=1.31\times2.80=3.668\approx4$, which matches the data.

Step3: Test option B with $x=2$

Substitute $x=2$ into $y=1.31(2.80^x)$:
$y=1.31\times2.80^2=1.31\times7.84=10.2704$, close to the data value 9.

Step4: Test option D with $x=1$

Substitute $x=1$ into $y=2.80(1.31^x)$:
$y=2.80\times1.31^1=3.668\approx4$, matches, but test $x=2$:
$y=2.80\times1.31^2=2.80\times1.7161=4.80508$, which is not close to 9.

Step5: Verify option B with $x=3$

Substitute $x=3$ into $y=1.31(2.80^x)$:
$y=1.31\times2.80^3=1.31\times21.952=28.75712$, close to 27, which aligns with the data's exponential growth trend.

Answer:

B. $y = 1.31(2.80^x)$