[tex]\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad
\begin{cases}
A=\textit{accumulated amount}\to &\$4600\\
P=\textit{original amount deposited}\\
r=rate\to 6.2\%\to \frac{6.2}{100}\to &0.062\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, so once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
4600=P\left(1+\frac{0.062}{1}\right)^{1\cdot 2}[/tex]
solve for P
