Respuesta :
[tex]~~~~~~ \underset{annually}{\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 &\$6500\\ r=rate\to 9\%\to \frac{9}{100}\dotfill &0.09\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &30 \end{cases}[/tex]
[tex]A=6500\left(1+\frac{0.09}{1}\right)^{1\cdot 30}\implies A=6500(1.09)^{30}\implies A\approx 86239.91 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \underset{monthly}{\textit{Compound Interest Earned Amount}}[/tex]
[tex]A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 9\%\to \frac{9}{100}\dotfill &0.09\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &30 \end{cases} \\\\\\ A=6500\left(1+\frac{0.09}{12}\right)^{12\cdot 30}\implies A=6500(1.0075)^{360}\implies A\approx 95748.74[/tex]
