Answer :
To solve this problem, let's use the formula for compound interest:
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount of money accumulated after [tex]\( t \)[/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 interest is compounded per year.
- [tex]\( t \)[/tex] is the number of years the money is invested for.
### a) Finding the function for the amount to which the investment grows after [tex]\( t \)[/tex] years:
Given:
- [tex]\( P = \$88,000 \)[/tex]
- [tex]\( r = 3.5\% = 0.035 \)[/tex]
- [tex]\( n = 4 \)[/tex] (since the interest is compounded quarterly)
Substitute [tex]\( P \)[/tex], [tex]\( r \)[/tex], and [tex]\( n \)[/tex] into the compound interest formula:
[tex]\[ A(t) = 88000 \left(1 + \frac{0.035}{4}\right)^{4t} \][/tex]
Simplify the expression inside the parentheses:
[tex]\[ A(t) = 88000 \left(1 + 0.00875\right)^{4t} \][/tex]
[tex]\[ A(t) = 88000 \left(1.00875\right)^{4t} \][/tex]
Thus, the function for the amount to which the investment grows after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 88000 \cdot 1.00875^{4t} \][/tex]
### b) Finding the amount of money in the account at [tex]\( t = 0, 2, 7, 10 \)[/tex] years:
Now, we will calculate [tex]\( A(t) \)[/tex] for [tex]\( t = 0, 2, 7, 10 \)[/tex]:
1. When [tex]\( t = 0 \)[/tex]:
[tex]\[ A(0) = 88000 \cdot 1.00875^{4 \cdot 0} \][/tex]
[tex]\[ A(0) = 88000 \cdot 1 = 88000 \][/tex]
So, the amount at [tex]\( t = 0 \)[/tex] years is [tex]\( \$88,000 \)[/tex].
2. When [tex]\( t = 2 \)[/tex]:
[tex]\[ A(2) = 88000 \cdot 1.00875^{4 \cdot 2} \][/tex]
[tex]\[ A(2) = 88000 \cdot 1.00875^8 \approx 94351.9877376592 \][/tex]
So, the amount at [tex]\( t = 2 \)[/tex] years is approximately [tex]\( \$94,351.99 \)[/tex].
3. When [tex]\( t = 7 \)[/tex]:
[tex]\[ A(7) = 88000 \cdot 1.00875^{4 \cdot 7} \][/tex]
[tex]\[ A(7) = 88000 \cdot 1.00875^{28} \approx 112310.926152495 \][/tex]
So, the amount at [tex]\( t = 7 \)[/tex] years is approximately [tex]\( \$112,310.93 \)[/tex].
4. When [tex]\( t = 10 \)[/tex]:
[tex]\[ A(10) = 88000 \cdot 1.00875^{4 \cdot 10} \][/tex]
[tex]\[ A(10) = 88000 \cdot 1.00875^{40} \approx 124687.977737940 \][/tex]
So, the amount at [tex]\( t = 10 \)[/tex] years is approximately [tex]\( \$124,687.98 \)[/tex].
### Summary:
- The function for the amount to which the investment grows after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 88000 \cdot 1.00875^{4t} \][/tex]
- The amounts in the account at different times are:
- [tex]\( A(0) = \$88,000 \)[/tex]
- [tex]\( A(2) \approx \$94,351.99 \)[/tex]
- [tex]\( A(7) \approx \$112,310.93 \)[/tex]
- [tex]\( A(10) \approx \$124,687.98 \)[/tex]
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A(t) \)[/tex] is the amount of money accumulated after [tex]\( t \)[/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 interest is compounded per year.
- [tex]\( t \)[/tex] is the number of years the money is invested for.
### a) Finding the function for the amount to which the investment grows after [tex]\( t \)[/tex] years:
Given:
- [tex]\( P = \$88,000 \)[/tex]
- [tex]\( r = 3.5\% = 0.035 \)[/tex]
- [tex]\( n = 4 \)[/tex] (since the interest is compounded quarterly)
Substitute [tex]\( P \)[/tex], [tex]\( r \)[/tex], and [tex]\( n \)[/tex] into the compound interest formula:
[tex]\[ A(t) = 88000 \left(1 + \frac{0.035}{4}\right)^{4t} \][/tex]
Simplify the expression inside the parentheses:
[tex]\[ A(t) = 88000 \left(1 + 0.00875\right)^{4t} \][/tex]
[tex]\[ A(t) = 88000 \left(1.00875\right)^{4t} \][/tex]
Thus, the function for the amount to which the investment grows after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 88000 \cdot 1.00875^{4t} \][/tex]
### b) Finding the amount of money in the account at [tex]\( t = 0, 2, 7, 10 \)[/tex] years:
Now, we will calculate [tex]\( A(t) \)[/tex] for [tex]\( t = 0, 2, 7, 10 \)[/tex]:
1. When [tex]\( t = 0 \)[/tex]:
[tex]\[ A(0) = 88000 \cdot 1.00875^{4 \cdot 0} \][/tex]
[tex]\[ A(0) = 88000 \cdot 1 = 88000 \][/tex]
So, the amount at [tex]\( t = 0 \)[/tex] years is [tex]\( \$88,000 \)[/tex].
2. When [tex]\( t = 2 \)[/tex]:
[tex]\[ A(2) = 88000 \cdot 1.00875^{4 \cdot 2} \][/tex]
[tex]\[ A(2) = 88000 \cdot 1.00875^8 \approx 94351.9877376592 \][/tex]
So, the amount at [tex]\( t = 2 \)[/tex] years is approximately [tex]\( \$94,351.99 \)[/tex].
3. When [tex]\( t = 7 \)[/tex]:
[tex]\[ A(7) = 88000 \cdot 1.00875^{4 \cdot 7} \][/tex]
[tex]\[ A(7) = 88000 \cdot 1.00875^{28} \approx 112310.926152495 \][/tex]
So, the amount at [tex]\( t = 7 \)[/tex] years is approximately [tex]\( \$112,310.93 \)[/tex].
4. When [tex]\( t = 10 \)[/tex]:
[tex]\[ A(10) = 88000 \cdot 1.00875^{4 \cdot 10} \][/tex]
[tex]\[ A(10) = 88000 \cdot 1.00875^{40} \approx 124687.977737940 \][/tex]
So, the amount at [tex]\( t = 10 \)[/tex] years is approximately [tex]\( \$124,687.98 \)[/tex].
### Summary:
- The function for the amount to which the investment grows after [tex]\( t \)[/tex] years is:
[tex]\[ A(t) = 88000 \cdot 1.00875^{4t} \][/tex]
- The amounts in the account at different times are:
- [tex]\( A(0) = \$88,000 \)[/tex]
- [tex]\( A(2) \approx \$94,351.99 \)[/tex]
- [tex]\( A(7) \approx \$112,310.93 \)[/tex]
- [tex]\( A(10) \approx \$124,687.98 \)[/tex]
