Question
2200 dollars is placed in an account with an annual interest rate of 7.25%. How much will be in the account after 29 years, to the nearest cent?
Answers
GPT-4o mini
To calculate the amount in the account after 29 years with an initial deposit of \( P = 2200 \) dollars, an annual interest rate of \( r = 7.25\% = 0.0725 \), and compounding interest, we can use the formula for compound interest:
\[
A = P(1 + r)^t
\]
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).
- \( t \) is the number of years the money is invested or borrowed.
Plugging in the values:
\[
A = 2200(1 + 0.0725)^{29}
\]
Calculating \( (1 + 0.0725)^{29} \):
\[
1 + 0.0725 = 1.0725
\]
\[
(1.0725)^{29} \approx 6.562164
\] (using a calculator for the exponentiation)
Now, substituting this back into the equation for \( A \):
\[
A \approx 2200 \times 6.562164 \approx 14435.85
\]
Thus, the amount in the account after 29 years will be approximately:
\[
\boxed{14435.85}
\]
to the nearest cent.
\[
A = P(1 + r)^t
\]
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).
- \( t \) is the number of years the money is invested or borrowed.
Plugging in the values:
\[
A = 2200(1 + 0.0725)^{29}
\]
Calculating \( (1 + 0.0725)^{29} \):
\[
1 + 0.0725 = 1.0725
\]
\[
(1.0725)^{29} \approx 6.562164
\] (using a calculator for the exponentiation)
Now, substituting this back into the equation for \( A \):
\[
A \approx 2200 \times 6.562164 \approx 14435.85
\]
Thus, the amount in the account after 29 years will be approximately:
\[
\boxed{14435.85}
\]
to the nearest cent.