To solve this problem, let's proceed with the steps one by one, using the given formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt}, \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (initial deposit),
- [tex]\( r \)[/tex] is the annual interest rate,
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year,
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( A \)[/tex] is the amount of money in the account after [tex]\( t \)[/tex] years.
### Part a: Find an equation that gives the amount of money in the account after [tex]\( t \)[/tex] years.
1. Identify the given variables:
- Principal amount ([tex]\( P \)[/tex]): [tex]\( \$1800 \)[/tex]
- Annual interest rate ([tex]\( r \)[/tex]): [tex]\( 5\% \)[/tex], or [tex]\( 0.05 \)[/tex] as a decimal
- Number of times the interest is compounded per year ([tex]\( n \)[/tex]): [tex]\( 4 \)[/tex] (quarterly)
2. Substitute these values into the formula:
[tex]\[ A(t) = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
So the equation that gives the amount of money in the account after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 1800 \left(1 + \frac{0.05}{4}\right)^{4t} \][/tex]
### Part b: Find the amount of money in the account after 8 years.
1. Substitute [tex]\( t = 8 \)[/tex] into the equation:
[tex]\[ A(8) = 1800 \left(1 + \frac{0.05}{4}\right)^{4 \times 8} \][/tex]
2. Simplify inside the parentheses:
[tex]\[ A(8) = 1800 \left(1 + \frac{0.05}{4}\right)^{32} \][/tex]
[tex]\[ A(8) = 1800 \left(1 + 0.0125\right)^{32} \][/tex]
[tex]\[ A(8) = 1800 \left(1.0125\right)^{32} \][/tex]
3. Calculate the value of [tex]\((1.0125)^{32}\)[/tex]:
[tex]\[ (1.0125)^{32} \approx 1.487019397 \][/tex]
4. Multiply this by 1800 to find [tex]\( A(8) \)[/tex]:
[tex]\[ A(8) = 1800 \times 1.487019397 \approx 2678.6349154706854 \][/tex]
5. Round the final amount to 2 decimal places:
[tex]\[ A(8) \approx 2678.63 \][/tex]
Therefore, after 8 years, there will be [tex]\( \$2678.63 \)[/tex] in the account.
