Solution
Note: Compound Interest Formula
From the question
Principal = $1500
Rate = 4.5% = 0.045
Time = 25 years
When n = 1
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \\ A=1500(1+\frac{0.045}{1})^{25} \\ \\ A=4508.15\text{ dollars \lparen to the nearest cent\rparen} \end{gathered}[/tex]
When n = 4
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \\ A=1500(1+\frac{0.045}{4})^{100} \\ \\ A=4591.40\text{ dollars \lparen to the nearest cent\rparen} \end{gathered}[/tex]
When n = 12
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \\ A=1500(1+\frac{0.045}{12})^{25\times12} \\ \\ A=4610.61\text{ dollars \lparen to the nearest cent\rparen} \end{gathered}[/tex]
When n = 365
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \\ A=1500(1+\frac{0.045}{365})^{25\times365} \\ \\ A=4620.00\text{ dollars \lparen to the nearest cent\rparen} \end{gathered}[/tex]
For compounded continuously
[tex]\begin{gathered} A=Pe^{rt} \\ A=1500e^{(0.045\times25)} \\ \\ A=4620.33\text{ dollars \lparen to the nearest cent\rparen} \end{gathered}[/tex]
