To find the final value of an amount subject to compound interest, we can use the compound interest formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after \( n \) years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (in decimal form).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for, in years.
Given:
- The principal \( P = $1300 \)
- The annual interest rate \( r = 14.5\% \) or \( 0.145 \) as a decimal.
- Interest is compounded monthly, so \( n = 12 \) times per year.
- The time \( t = 4.5 \) years.
We can plug these values into the formula:
\[ A = 1300 \left(1 + \frac{0.145}{12}\right)^{(12 \cdot 4.5)} \]
First, we will calculate the rate per period \( \frac{0.145}{12} \):
\[ \text{Rate per period} = \frac{0.145}{12} \approx 0.01208333... \]
Substitute this into the formula:
\[ A = 1300 \left(1 + 0.01208333\right)^{(12 \cdot 4.5)} \]
Now, let's calculate the total number of periods:
\[ n \cdot t = 12 \cdot 4.5 = 54 \]
And raise the expression inside the parentheses to the 54th power:
\[ A = 1300 \left(1.01208333\right)^{54} \]
To compute the accumulated value \( A \), you would typically use a calculator:
\[ A \approx 1300 \times (1.01208333)^{54} \]
\[ A \approx 1300 \times 1.90642485... \]
\[ A \approx 2478.3523075 \]
Finally, we round this to the nearest cent:
\[ A \approx \$2478.35 \]
Thus, the final value of the amount after 4.5 years is approximately $2478.35.
