[tex]\bf \qquad \textit{Simple Interest Earned Amount}\\\\
A=P(1+rt)\qquad
\begin{cases}
A=\textit{accumulated amount}\to &\$2419.60\\
P=\textit{original amount deposited}\\
r=rate\to 5.2\%\to \frac{5.2}{100}\to &0.052\\
t=years\to &1
\end{cases}
\\\\\\
2419.60=P(1+0.052\cdot 1)[/tex]
solve for P
