What's the balance (to the nearest cent) after 11 yr if you invest $700 at a nominal annual rate of 4.5% if:

Given that the investment money is $700. The nominal annual interest rate is 4.5% and the time period is 11 years.
We have to find the amount at given time period.
a)
The formula when the interest is compounded annually is:
[tex]A=P(1+r)^t[/tex]Substitute the given values in the formula:
[tex]\begin{gathered} A=700(1+0.045)^{11} \\ =700(1.623) \\ =1136.1 \end{gathered}[/tex]Thus, the answer is $1136.1.
b)
The formula of amount when the interest is compounded weekly is:
[tex]A=P(1+\frac{7r}{365})^{\frac{365}{7}t}[/tex]Substitute the given values in the formula:
[tex]\begin{gathered} A=700(1+\frac{7\times0.045}{365})^{\frac{365\times11}{7}} \\ =700(1+0.000863)^{573.57} \\ =700(1.000863)^{573.57} \\ =700(1.6401) \\ =1148.07 \end{gathered}[/tex]Thus, the answer is $1148.07.
c)
The formula of amount when the interest is compounded daily is:
[tex]A=P(1+\frac{r}{365})^{365t}[/tex]Substitute the given values in the formula:
[tex]\begin{gathered} A=700(1+\frac{0.045}{365})^{365\times11} \\ =700(1+0.0001232)^{4015} \\ =700(1.0001232)^{4015} \\ =700(1.6398) \\ =1147.86 \end{gathered}[/tex]Thus, the answer is $1147.86.
d)
The formula when the interest is compounded continuously is:
[tex]A=Pe^{rt}[/tex]substitute the given values in the formula:
[tex]\begin{gathered} A=700e^{(0.045\times11)} \\ =700e^{0.495} \\ =700(1.6404) \\ =1148.28 \end{gathered}[/tex]Thus, the answer is $1148.28.