Answer :
Let's work through the problem step-by-step.
## Part (a)
To find the function for the amount to which the investment grows after [tex]\( t \)[/tex] years, we use the compound interest formula:
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( P \)[/tex] is the initial principal (the initial investment)
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal)
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year
- [tex]\( t \)[/tex] is the number of years
- [tex]\( A(t) \)[/tex] is the amount after [tex]\( t \)[/tex] years
In this problem:
- The initial principal [tex]\( P \)[/tex] is \[tex]$88,000. - The annual interest rate \( r \) is \( 4.5\% \), which is \( 0.045 \) when expressed as a decimal. - The interest is compounded quarterly, so \( n = 4 \). Plugging these values into the formula, we get: \[ A(t) = 88000 \left(1 + \frac{0.045}{4}\right)^{4t} \] Simplify the expression inside the parentheses: \[ A(t) = 88000 \left(1 + 0.01125\right)^{4t} \] \[ A(t) = 88000 \left(1.01125\right)^{4t} \] So, the function for the amount to which the investment grows after \( t \) years is: \[ A(t) = 88000 \left(1.01125\right)^{4t} \] ## Part (b) Now, we'll calculate the amount of money in the account at \( t = 0, 4, 5, \) and \( 10 \) years using the function we derived above. ### At \( t = 0 \) years \[ A(0) = 88000 \left(1.01125\right)^{4 \cdot 0} \] \[ A(0) = 88000 \left(1.01125\right)^0 \] \[ A(0) = 88000 \cdot 1 \] \[ A(0) = 88000 \] So, at \( t = 0 \) years, the amount is \$[/tex]88,000.
### At [tex]\( t = 4 \)[/tex] years
[tex]\[ A(4) = 88000 \left(1.01125\right)^{4 \cdot 4} \][/tex]
[tex]\[ A(4) = 88000 \left(1.01125\right)^{16} \][/tex]
Using a precise calculation method, we determine:
[tex]\[ A(4) \approx 105249.13 \][/tex]
So, at [tex]\( t = 4 \)[/tex] years, the amount is approximately \[tex]$105,249.30. ### At \( t = 5 \) years \[ A(5) = 88000 \left(1.01125\right)^{4 \cdot 5} \] \[ A(5) = 88000 \left(1.01125\right)^{20} \] Using a precise calculation method, we determine: \[ A(5) \approx 110066.05 \] So, at \( t = 5 \) years, the amount is approximately \$[/tex]110,066.05.
### At [tex]\( t = 10 \)[/tex] years
[tex]\[ A(10) = 88000 \left(1.01125\right)^{4 \cdot 10} \][/tex]
[tex]\[ A(10) = 88000 \left(1.01125\right)^{40} \][/tex]
Using a precise calculation method, we determine:
[tex]\[ A(10) \approx 137665.16 \][/tex]
So, at [tex]\( t = 10 \)[/tex] years, the amount is approximately \[tex]$137,665.16. Therefore, the amounts in the account at \( t = 0, 4, 5, \) and \( 10 \) years are as follows: - At \( t = 0 \): \$[/tex]88,000
- At [tex]\( t = 4 \)[/tex]: \[tex]$105,249.30 - At \( t = 5 \): \$[/tex]110,066.05
- At [tex]\( t = 10 \)[/tex]: \[tex]$137,665.16 Hence, the final answer to part (a) is: \[ A(t) = 88000 \left(1.01125\right)^{4t} \] And the final answer to part (b) is: - At \( t = 0 \): \$[/tex]88,000
- At [tex]\( t = 4 \)[/tex]: \[tex]$105,249.30 - At \( t = 5 \): \$[/tex]110,066.05
- At [tex]\( t = 10 \)[/tex]: \$137,665.16
## Part (a)
To find the function for the amount to which the investment grows after [tex]\( t \)[/tex] years, we use the compound interest formula:
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( P \)[/tex] is the initial principal (the initial investment)
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal)
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year
- [tex]\( t \)[/tex] is the number of years
- [tex]\( A(t) \)[/tex] is the amount after [tex]\( t \)[/tex] years
In this problem:
- The initial principal [tex]\( P \)[/tex] is \[tex]$88,000. - The annual interest rate \( r \) is \( 4.5\% \), which is \( 0.045 \) when expressed as a decimal. - The interest is compounded quarterly, so \( n = 4 \). Plugging these values into the formula, we get: \[ A(t) = 88000 \left(1 + \frac{0.045}{4}\right)^{4t} \] Simplify the expression inside the parentheses: \[ A(t) = 88000 \left(1 + 0.01125\right)^{4t} \] \[ A(t) = 88000 \left(1.01125\right)^{4t} \] So, the function for the amount to which the investment grows after \( t \) years is: \[ A(t) = 88000 \left(1.01125\right)^{4t} \] ## Part (b) Now, we'll calculate the amount of money in the account at \( t = 0, 4, 5, \) and \( 10 \) years using the function we derived above. ### At \( t = 0 \) years \[ A(0) = 88000 \left(1.01125\right)^{4 \cdot 0} \] \[ A(0) = 88000 \left(1.01125\right)^0 \] \[ A(0) = 88000 \cdot 1 \] \[ A(0) = 88000 \] So, at \( t = 0 \) years, the amount is \$[/tex]88,000.
### At [tex]\( t = 4 \)[/tex] years
[tex]\[ A(4) = 88000 \left(1.01125\right)^{4 \cdot 4} \][/tex]
[tex]\[ A(4) = 88000 \left(1.01125\right)^{16} \][/tex]
Using a precise calculation method, we determine:
[tex]\[ A(4) \approx 105249.13 \][/tex]
So, at [tex]\( t = 4 \)[/tex] years, the amount is approximately \[tex]$105,249.30. ### At \( t = 5 \) years \[ A(5) = 88000 \left(1.01125\right)^{4 \cdot 5} \] \[ A(5) = 88000 \left(1.01125\right)^{20} \] Using a precise calculation method, we determine: \[ A(5) \approx 110066.05 \] So, at \( t = 5 \) years, the amount is approximately \$[/tex]110,066.05.
### At [tex]\( t = 10 \)[/tex] years
[tex]\[ A(10) = 88000 \left(1.01125\right)^{4 \cdot 10} \][/tex]
[tex]\[ A(10) = 88000 \left(1.01125\right)^{40} \][/tex]
Using a precise calculation method, we determine:
[tex]\[ A(10) \approx 137665.16 \][/tex]
So, at [tex]\( t = 10 \)[/tex] years, the amount is approximately \[tex]$137,665.16. Therefore, the amounts in the account at \( t = 0, 4, 5, \) and \( 10 \) years are as follows: - At \( t = 0 \): \$[/tex]88,000
- At [tex]\( t = 4 \)[/tex]: \[tex]$105,249.30 - At \( t = 5 \): \$[/tex]110,066.05
- At [tex]\( t = 10 \)[/tex]: \[tex]$137,665.16 Hence, the final answer to part (a) is: \[ A(t) = 88000 \left(1.01125\right)^{4t} \] And the final answer to part (b) is: - At \( t = 0 \): \$[/tex]88,000
- At [tex]\( t = 4 \)[/tex]: \[tex]$105,249.30 - At \( t = 5 \): \$[/tex]110,066.05
- At [tex]\( t = 10 \)[/tex]: \$137,665.16
