Answer :
To determine how much money Avery will have in the account after 15 years, we need to use the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( n \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Given:
- [tex]\( P = 8300 \)[/tex] (the initial investment)
- [tex]\( r = 0.022 \)[/tex] (the annual interest rate in decimal form)
- [tex]\( n = 4 \)[/tex] (quarterly compounding means 4 times per year)
- [tex]\( t = 15 \)[/tex] (the investment period in years)
Now we can plug these values into the formula:
[tex]\[ A = 8300 \left(1 + \frac{0.022}{4}\right)^{4 \times 15} \][/tex]
First, calculate the term inside the parentheses:
[tex]\[ \frac{0.022}{4} = 0.0055 \][/tex]
So the formula becomes:
[tex]\[ A = 8300 \left(1 + 0.0055\right)^{60} \][/tex]
Next, compute [tex]\( 1 + 0.0055 \)[/tex]:
[tex]\[ 1 + 0.0055 = 1.0055 \][/tex]
Then, raise [tex]\( 1.0055 \)[/tex] to the power of 60 (since [tex]\( n \times t = 4 \times 15 = 60 \)[/tex]):
[tex]\[ 1.0055^{60} \approx 1.34935 \][/tex]
Now multiply this result by the principal amount, [tex]\( P \)[/tex]:
[tex]\[ A = 8300 \times 1.34935 \approx 11,199.605 \][/tex]
Therefore, the amount accumulated in the account after 15 years, rounded to the nearest cent, is:
[tex]\[ A \approx 11,199.61 \][/tex]
So, Avery would have $11,199.61 in the account after 15 years.
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( n \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Given:
- [tex]\( P = 8300 \)[/tex] (the initial investment)
- [tex]\( r = 0.022 \)[/tex] (the annual interest rate in decimal form)
- [tex]\( n = 4 \)[/tex] (quarterly compounding means 4 times per year)
- [tex]\( t = 15 \)[/tex] (the investment period in years)
Now we can plug these values into the formula:
[tex]\[ A = 8300 \left(1 + \frac{0.022}{4}\right)^{4 \times 15} \][/tex]
First, calculate the term inside the parentheses:
[tex]\[ \frac{0.022}{4} = 0.0055 \][/tex]
So the formula becomes:
[tex]\[ A = 8300 \left(1 + 0.0055\right)^{60} \][/tex]
Next, compute [tex]\( 1 + 0.0055 \)[/tex]:
[tex]\[ 1 + 0.0055 = 1.0055 \][/tex]
Then, raise [tex]\( 1.0055 \)[/tex] to the power of 60 (since [tex]\( n \times t = 4 \times 15 = 60 \)[/tex]):
[tex]\[ 1.0055^{60} \approx 1.34935 \][/tex]
Now multiply this result by the principal amount, [tex]\( P \)[/tex]:
[tex]\[ A = 8300 \times 1.34935 \approx 11,199.605 \][/tex]
Therefore, the amount accumulated in the account after 15 years, rounded to the nearest cent, is:
[tex]\[ A \approx 11,199.61 \][/tex]
So, Avery would have $11,199.61 in the account after 15 years.
