[tex]\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad
\begin{cases}
A=\textit{compounded amount}\\
P=\textit{original amount deposited}\to &\$1320\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, meaning once}
\end{array}\to &1\\
t=years\to &5
\end{cases}[/tex]
he'll be earning in interest, the difference from A to P, or A - P
