Respuesta :
[tex]\bf \qquad \textit{Simple Interest Earned Amount}\\\\
A=P(1+rt)\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$3000\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
t=years\to &4
\end{cases}
\\\\\\
A=3000(1+0.05\cdot 4)\implies \boxed{A=3600}\\\\
-------------------------------\\\\[/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 &\$3000\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &4 \end{cases}[/tex]
[tex]\bf A=3000\left(1+\frac{0.05}{1}\right)^{1\cdot 4}\implies A=3000(1.05)^4\implies \boxed{A=3646.51875}\\\\ -------------------------------\\\\ \stackrel{\textit{compounded interest}}{3646.51875}~~-~~\stackrel{\textit{simple interest}}{3600}[/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 &\$3000\\ r=rate\to 5\%\to \frac{5}{100}\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &4 \end{cases}[/tex]
[tex]\bf A=3000\left(1+\frac{0.05}{1}\right)^{1\cdot 4}\implies A=3000(1.05)^4\implies \boxed{A=3646.51875}\\\\ -------------------------------\\\\ \stackrel{\textit{compounded interest}}{3646.51875}~~-~~\stackrel{\textit{simple interest}}{3600}[/tex]
