Respuesta :
[tex]\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
\stackrel{Bank~A}{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 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\
t=years\to &1
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.06}{2}\right)^{2\cdot 1}\implies A=1000(1.03)^2\implies \boxed{A=1060.9}[/tex]
[tex]\bf -------------------------------\\\\ \qquad \textit{Compound Interest Earned Amount} \\\\ \stackrel{Bank~B}{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 6\%\to \frac{6}{100}\to &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve}\\ \end{array}\to &12\\ t=years\to &1 \end{cases} [/tex]
[tex]\bf A=1000\left(1+\frac{0.06}{12}\right)^{2\cdot 1}\implies A=1000(1.005)^{12}~~\approx~~ \boxed{A=1061.67781}\\\\ -------------------------------\\\\ thus\qquad \qquad 1061.67781~-~1060.9[/tex]
[tex]\bf -------------------------------\\\\ \qquad \textit{Compound Interest Earned Amount} \\\\ \stackrel{Bank~B}{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 6\%\to \frac{6}{100}\to &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve}\\ \end{array}\to &12\\ t=years\to &1 \end{cases} [/tex]
[tex]\bf A=1000\left(1+\frac{0.06}{12}\right)^{2\cdot 1}\implies A=1000(1.005)^{12}~~\approx~~ \boxed{A=1061.67781}\\\\ -------------------------------\\\\ thus\qquad \qquad 1061.67781~-~1060.9[/tex]