Answer
a. $A=110e^{0.1013t}$ million.
b. $13\ years$, or in year 2013.
Work Step by Step
a. Let year 2000 corresponding to $t=0$, we have $A_0=110$. In year 2010, $t=10$ and $303=110e^{10k}$ which gives $k=\frac{ln(303/110)}{10}\approx0.1013$ and the model equation becomes $A=110e^{0.1013t}$ million.
b. Let $A=400$; we have $400=110e^{0.1013t}$ and thus $t=\frac{ln(400/110)}{0.1013}\approx13\ years$ or in year 2013.
