Respuesta :
[tex]\bf ~~~~~~ \textit{Compounding Continuously Interest Earned Amount}\\\\
A=Pe^{rt}\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$1000\\
r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\
t=years\to &5
\end{cases}
\\\\\\
A=1000e^{0.037\cdot 5}\implies A=1000e^{0.185}\\\\
-------------------------------\\\\[/tex]
[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}\to &\$1000\\ r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\to &4\\ t=years\to &5 \end{cases} \\\\\\ A=1000\left(1+\frac{0.037}{4}\right)^{4\cdot 5}\implies A=1000(1.00925)^{20}[/tex]
compare both amounts.
[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}\to &\$1000\\ r=rate\to 3.7\%\to \frac{3.7}{100}\to &0.037\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\to &4\\ t=years\to &5 \end{cases} \\\\\\ A=1000\left(1+\frac{0.037}{4}\right)^{4\cdot 5}\implies A=1000(1.00925)^{20}[/tex]
compare both amounts.
