Answer
a) $\$853.29$
b) $1992$ and $2000$
Work Step by Step
Given \begin{equation}
B(t)=\frac{-470001 t^2+4110992 t+14032612}{-469.4 t^2+3745 t+19774}.
\end{equation} a) The year $1995$ corresponds to $t= 5$ since $1990$. Find $B(5)$: \begin{equation}
\begin{aligned}
B(5)&= \frac{-470001\cdot 5^2+4110992 \cdot 5+14032612}{-469.4 \cdot 5^2+3745\cdot 5+19774}\\
&=\frac{22837547}{26764}\\
&\approx 853.29
\end{aligned}
\end{equation} The average benefit for a person participating in the U.S. food stamp program in 1995 was about $\$853.29$.
b) Set $B(t) = 800$ and graph the right and left hand side functions in the same window.
\begin{equation}
\begin{aligned}
B(t) & = 800\\
\frac{-470001 t^2+4110992 t+14032612}{-469.4 t^2+3745 t+19774}&= 800\\
-470001 t^2+4110992 t+14032612&=800\left(-469.4 t^2+3745 t+19774\right).
\end{aligned}
\end{equation} Let \begin{equation}
\begin{aligned}
f(x)& =-470001 t^2+4110992 t+14032612\\
g(x)&= 800\left(-469.4 t^2+3745 t+19774\right).
\end{aligned}
\end{equation} The point of intersection between the two functions occurs at about $t= 2$ and about $t= 10$. Hence, the average benefit for a person participating in the U.S. food stamp program was approximately $\$800$ in $1992$ and again in $2000$.
