[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 &\$4400\\
P=\textit{original amount deposited}\\
r=rate\to 4.8\%\to \frac{4.8}{100}\to &0.048\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
4400=P\left(1+\frac{0.048}{1}\right)^{1\cdot 2}\implies \cfrac{4000}{\left( 1+0.048 \right)^2}=P[/tex]
