Answer
(a) $n(t)=1500e^{0.1515t}$
(b) $7940$
Work Step by Step
(a) We can obtain from the graph that $n_0=1500$ with the data point given on the graph,
the function becomes $n(5)=1500e^{5r}=3200$, we obtain $r=ln(\frac{3200}{1500})/5=0.1515$ and
the model function becomes $n(t)=1500e^{0.1515t}$
(b) Year 2020 corresponds to $t=2020-2009=11$, we have $n(11)=1500e^{0.1515\times11}\approx7940$