Question
Emily puts 5,619 into account to use for school expenses the account earns 4.04% interest compounded quarterly how much will be in the account after 10 years use the formula a = p * 1 + are over in to the power of NT where a is the balance he is the principal or is the interest rate expressed as a decimal and is the number of times per year that the interest is compounded and t is the time in years around your answer to the nearest
Answers
GPT-4o mini
To calculate the amount in the account after 10 years, we can use the formula for compound interest:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
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 (decimal).
- \(n\) is the number of times that interest is compounded per year.
- \(t\) is the number of years the money is invested or borrowed.
Given:
- \(P = 5619\)
- \(r = 4.04\% = 0.0404\) (as a decimal)
- \(n = 4\) (quarterly compounding)
- \(t = 10\)
Now we can plug these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \times 10}
\]
Calculating \(\frac{0.0404}{4}\):
\[
\frac{0.0404}{4} = 0.0101
\]
Now plug this into the formula:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
This becomes:
\[
A = 5619 (1.0101)^{40}
\]
Now we need to calculate \((1.0101)^{40}\):
\[
(1.0101)^{40} \approx 1.488864
\]
Now multiply this by the principal:
\[
A \approx 5619 \times 1.488864 \approx 8366.40
\]
Rounding to the nearest dollar:
\[
A \approx 8366
\]
Thus, the amount in the account after 10 years will be approximately **$8,366**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
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 (decimal).
- \(n\) is the number of times that interest is compounded per year.
- \(t\) is the number of years the money is invested or borrowed.
Given:
- \(P = 5619\)
- \(r = 4.04\% = 0.0404\) (as a decimal)
- \(n = 4\) (quarterly compounding)
- \(t = 10\)
Now we can plug these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \times 10}
\]
Calculating \(\frac{0.0404}{4}\):
\[
\frac{0.0404}{4} = 0.0101
\]
Now plug this into the formula:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
This becomes:
\[
A = 5619 (1.0101)^{40}
\]
Now we need to calculate \((1.0101)^{40}\):
\[
(1.0101)^{40} \approx 1.488864
\]
Now multiply this by the principal:
\[
A \approx 5619 \times 1.488864 \approx 8366.40
\]
Rounding to the nearest dollar:
\[
A \approx 8366
\]
Thus, the amount in the account after 10 years will be approximately **$8,366**.