[tex]\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4700\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &12
\end{cases}
\\\\\\
A=4700\left(1+\frac{0.06}{1}\right)^{1\cdot 12}\implies A=4700(1+0.06)^{12}[/tex]
