Respuesta :
[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 &\$3100\\ r=rate\to 6.5\%\to \frac{6.5}{100}\dotfill &0.065\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &12 \end{cases} \\\\\\ A=3100\left(1+\frac{0.065}{1}\right)^{1\cdot 12}\implies A=3100(1.065)^{12}\implies A\approx 6600.198[/tex]
