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 cent
Answers
GPT-4o mini
To calculate the amount in Emily's account after 10 years, we can use the formula for compound interest:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount of money).
- \( r \) = the annual interest rate (decimal).
- \( n \) = the number of times that interest is compounded per year.
- \( t \) = the number of years the money is invested for.
Given:
- \( P = 5619 \)
- \( r = 4.04\% = 0.0404 \) (as a decimal)
- \( n = 4 \) (since the interest is compounded quarterly)
- \( t = 10 \)
Now plugging in these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \times 10}
\]
First, calculate \( \frac{0.0404}{4} \):
\[
\frac{0.0404}{4} = 0.0101
\]
Now, substitute this value into the formula:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
\[
A = 5619 \left(1.0101\right)^{40}
\]
Next, calculate \( (1.0101)^{40} \):
\[
(1.0101)^{40} \approx 1.448886
\]
Now plug this back into the equation for \( A \):
\[
A = 5619 \times 1.448886 \approx 8151.64
\]
Thus, the total amount in the account after 10 years is approximately:
\[
\boxed{8151.64}
\]
rounded to the nearest cent.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount of money).
- \( r \) = the annual interest rate (decimal).
- \( n \) = the number of times that interest is compounded per year.
- \( t \) = the number of years the money is invested for.
Given:
- \( P = 5619 \)
- \( r = 4.04\% = 0.0404 \) (as a decimal)
- \( n = 4 \) (since the interest is compounded quarterly)
- \( t = 10 \)
Now plugging in these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \times 10}
\]
First, calculate \( \frac{0.0404}{4} \):
\[
\frac{0.0404}{4} = 0.0101
\]
Now, substitute this value into the formula:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
\[
A = 5619 \left(1.0101\right)^{40}
\]
Next, calculate \( (1.0101)^{40} \):
\[
(1.0101)^{40} \approx 1.448886
\]
Now plug this back into the equation for \( A \):
\[
A = 5619 \times 1.448886 \approx 8151.64
\]
Thus, the total amount in the account after 10 years is approximately:
\[
\boxed{8151.64}
\]
rounded to the nearest cent.