Answer :
To calculate the future value of $11,000 with a 2.25% interest rate compounded semiannually over 6 years, you can use the formula for compound interest:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the future value of the investment/loan, including interest
- \( P \) is the principal investment amount (the initial deposit or loan amount) ($11,000 in this case)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times that interest is compounded per unit \( t \) (6 years in this case)
- \( t \) is the time the money is invested for in years
Plugging in the values:
\[ A = 11000 \times \left(1 + \frac{0.0225}{2}\right)^{(2 \times 6)} \]
\[ A = 11000 \times \left(1 + 0.01125\right)^{12} \]
\[ A = 11000 \times (1.01125)^{12} \]
Now, calculate \( (1.01125)^{12} \):
\[ (1.01125)^{12} ≈ 1.14073 \]
Now, multiply:
\[ A ≈ 11000 \times 1.14073 \]
\[ A ≈ 12548.03 \]
So, the future value of $11,000 with a 2.25% interest rate compounded semiannually over 6 years is approximately $12,548.03.
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the future value of the investment/loan, including interest
- \( P \) is the principal investment amount (the initial deposit or loan amount) ($11,000 in this case)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times that interest is compounded per unit \( t \) (6 years in this case)
- \( t \) is the time the money is invested for in years
Plugging in the values:
\[ A = 11000 \times \left(1 + \frac{0.0225}{2}\right)^{(2 \times 6)} \]
\[ A = 11000 \times \left(1 + 0.01125\right)^{12} \]
\[ A = 11000 \times (1.01125)^{12} \]
Now, calculate \( (1.01125)^{12} \):
\[ (1.01125)^{12} ≈ 1.14073 \]
Now, multiply:
\[ A ≈ 11000 \times 1.14073 \]
\[ A ≈ 12548.03 \]
So, the future value of $11,000 with a 2.25% interest rate compounded semiannually over 6 years is approximately $12,548.03.
