[tex]\bf ~~~~~~ \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}\dotfill &\$10500\\ r=rate\to 2.99\%\to \frac{2.99}{100}\dotfill &0.0299\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=10500\left(1+\frac{0.0299}{1}\right)^{1\cdot 5}\implies A=10500(1.0299)^5\implies A\approx 12166.47[/tex]
