Respuesta :
[tex]\bf \qquad \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}\to &\$12000\\
r=rate\to 9\%\to \frac{9}{100}\to &0.09\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four times}
\end{array}\to &4\\
t=years\to &2
\end{cases}
\\\\\\
A=12000\left(1+\frac{0.09}{4}\right)^{4\cdot 2}[/tex]
[tex]\bf \qquad \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}\to &\$12000\\ r=rate\to 8.85\%\to \frac{8.85}{100}\to &0.0885\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve times} \end{array}\to &12\\ t=years\to &2 \end{cases} \\\\\\ A=12000\left(1+\frac{0.0885}{12}\right)^{12\cdot 2}[/tex]
[tex]\bf \qquad \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}\to &\$12000\\ r=rate\to 8.85\%\to \frac{8.85}{100}\to &0.0885\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve times} \end{array}\to &12\\ t=years\to &2 \end{cases} \\\\\\ A=12000\left(1+\frac{0.0885}{12}\right)^{12\cdot 2}[/tex]
