Answer
a) $N(t)= \frac{8215.1 t^2-23035.4 t+413525.6}{0.226 t^2+1.885 t+204.72}$
b) $\$15,314.21$.
c) $1990$
Work Step by Step
Given \begin{equation}
\begin{aligned}
P(t)&=0.226 t^2+1.885 t+204.72\\
D(t)&=8215.1 t^2-23035.4 t+413525.6
\end{aligned}
\end{equation} where $P(t)$ is the population of the of the United States in millions $t$ years since $1970$, and $D(t)$ is the national debt of the United States in millions of dollars $t$ years since $1970$. a) Let $N(t)$ represents the average national debt of the United States in millions of dollars $t$ years since $1970$. Then, we have $N(t) = \frac{D(t)}{P(t)}$. \begin{equation}
\begin{aligned}
N(t)&= \frac{8215.1 t^2-23035.4 t+413525.6}{0.226 t^2+1.885 t+204.72}.
\end{aligned}
\end{equation} b) There are $30$ years between $1970$ and $2000$. Set $t= 30$ and find $N(30)$: \begin{equation}
\begin{aligned}
N(t)&=\frac{8215.1 \cdot 30^2-23035.4 \cdot 30+413525.6}{0.226\cdot 30^2+1.885 \cdot 30+204.72} \\
&=15314.21.
\end{aligned}
\end{equation} The average amount of national debt per person in $2000$ was about $\$15,314.21$.
c) Find $t$ by setting $N(t) = 10000$: \begin{equation}
\begin{aligned}
N(t)&= 10000\\
\frac{8215.1 t^2-23035.4 t+413525.6}{0.226 t^2+1.885 t+204.72}&= 10000\\
8215.1 t^2-23035.4 t+413525.6=10000\left(0.226 t^2+1.885 t+204.72\right)\\
8215.1 t^2-23035.4 t+413525.6-10000\left(0.226 t^2+1.885 t+204.72\right)&= 0\\
8215.1 t^2-23035.4 t+413525.6-2260 t^2-18850 t-2047200&=0\\
5955.1 t^2-41885.4 t-1633674.4&=0.
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
t&=\frac{-(-418854) \pm \sqrt{(-418854)^2-4 \cdot 59551(-16336744)}}{2 \cdot 59551}\\
&= \frac{-(-418854) \pm \sqrt{418854^2+3891477767776}}{2 \cdot 59551}.
\end{aligned}
\end{equation} This gives $$
t=20.45 \ldots, t=-13.42.
$$ Use only the positive value of $t= 20.45$, which corresponds to the year $1990$. The average amount of national debt per person was $\$10,000$ in mid $1990$.
We could have also graphed $N(t)$ and $g(t) = 10000$. See the graph.
