Answer
a. $y=201.2(1.011)^x$ see explanations.
b. $y= 201.2 e^{0.0109x}$, $1.1\%$
Work Step by Step
a. Plot the data and use $y=ab^x$ to fit; we can get $a=201.2, b=1.011, r=0.999$, that is $y=201.2(1.011)^x$ as shown in the figure. As $r$ is close to $1$, it indicates a very good fit using the model.
b. We can rewrite the model as
$y=ab^x=ae^{ln(b^x)}=201.2 e^{ln((1.011)^x)}=201.2 e^{x\ ln1.011}=201.2 e^{0.0109x}$
The increasing percentage each year can be found as
$\frac{y(1)}{y(0)}-1=e^{0.0109}-1\approx0.011$ or $1.1\%$
