Answer :
To calculate the amount of money you will have after 10 years with a $8,900 deposit at a 2% interest rate compounded quarterly, you can use the formula for compound interest:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after \( t \) years, including interest.
- \( P \) is the principal amount (the initial deposit).
- \( r \) is the annual interest rate (in decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for, in years.
Given:
- \( P = $8,900 \)
- \( r = 0.02 \) (2% interest rate in decimal)
- \( n = 4 \) (compounded quarterly)
- \( t = 10 \) years
Substitute the values into the formula:
\[ A = 8900 \times \left(1 + \frac{0.02}{4}\right)^{4 \times 10} \]
\[ A = 8900 \times \left(1 + 0.005\right)^{40} \]
\[ A = 8900 \times (1.005)^{40} \]
\[ A = 8900 \times (1.221386) \]
\[ A \approx 10868.32 \]
So, you will have approximately $10,868.32 after 10 years with a $8,900 deposit at a 2% interest rate compounded quarterly.
Hope this helps!
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after \( t \) years, including interest.
- \( P \) is the principal amount (the initial deposit).
- \( r \) is the annual interest rate (in decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested for, in years.
Given:
- \( P = $8,900 \)
- \( r = 0.02 \) (2% interest rate in decimal)
- \( n = 4 \) (compounded quarterly)
- \( t = 10 \) years
Substitute the values into the formula:
\[ A = 8900 \times \left(1 + \frac{0.02}{4}\right)^{4 \times 10} \]
\[ A = 8900 \times \left(1 + 0.005\right)^{40} \]
\[ A = 8900 \times (1.005)^{40} \]
\[ A = 8900 \times (1.221386) \]
\[ A \approx 10868.32 \]
So, you will have approximately $10,868.32 after 10 years with a $8,900 deposit at a 2% interest rate compounded quarterly.
Hope this helps!
