Given
Part A
[tex]\begin{gathered} P=\text{ \$6,500} \\ r=3\text{ \%} \\ t=1 \end{gathered}[/tex]
Formula
[tex]A=P(1+\frac{r}{n})^{nt}[/tex][tex]\begin{gathered} A=6,500(1+\frac{0.03}{1})^{1\times1} \\ \\ A=6,500(1.03) \\ A=\text{ \$}6,695 \end{gathered}[/tex]
Part B
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \\ P=6500 \\ R=3\text{ \%} \\ t=2 \end{gathered}[/tex][tex]\begin{gathered} A=6500(1+\frac{0.03}{1})^{1\times2} \\ \\ A=6500(1.03)^2 \\ A=6500(1.0609) \\ A=\text{ \$}6895.85 \end{gathered}[/tex]
The final answer
