Question: Suppose that $88,000 is invested at 6% interest, compounded quarterly. a) Find the function for…

Suppose that $88,000 is invested at 6% interest, compounded quarterly.

a) Find the function for the amount to which the investment grows after t years. b) Graph the function. c) Find the amount of money in the account at t = 0, 3, 5, and 10 years. d) When will the amount of money in the account reach $300,000?

a) The function for the amount to which the investment grows after t years is A(t) = 88000(1.01625)^t (Simplify your answer. Type an expression using t as the variable.)

b) Graph the function. Choose the correct answer below. Use the viewing window [0, 50, 0, 500,000] with XscI = 5 and YscI = 50,000.

c) Find the amount of money in the account at t = 0, 3, 5, and 10 years.

The amount of money in the account at t = 0 years is $88,000. (Round to the nearest dollar as needed.)

The amount of money in the account at t = 3 years is $106,780. (Round to the nearest dollar as needed.)

The amount of money in the account at t = 5 years is $147,477. (Round to the nearest dollar as needed.)

The amount of money in the account at t = 10 years is $____. (Round to the nearest dollar as needed.)

Solution

To solve the given problem, we need to address each part step by step. The scenario involves $88,000 invested at an interest rate of \(6 \frac{1}{2}\%\), compounded quarterly. ### a) Find the function for the amount to which the investment grows after \( t \) years. The formula for compound interest is: \[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( P = 88000 \) is the principal amount, - \( r = 0.065 \) is the annual interest rate (6.5%), - \( n = 4 \) is the number of times interest applied per year (quarterly), - \( t \) is the time in years. Substitute the values into the formula: \[ A(t) = 88000 \left(1 + \frac{0.065}{4}\right)^{4t} \] \[ A(t) = 88000 \left(1 + 0.01625\right)^{4t} \] \[ A(t) = 88000 (1.01625)^{4t} \] ### b) Graph the function. Option B with a steadily increasing curve is the correct graph for this exponential growth function. ### c) Find the amount of money in the account at \( t = 0, 3, 5, \) and \( 10 \) years. 1. For \( t = 0 \): \[ A(0) = 88000 (1.01625)^{0} = 88000 \] 2. For \( t = 3 \): \[ A(3) = 88000 (1.01625)^{12} \] \[ A(3) \approx 106,780 \] 3. For \( t = 5 \): \[ A(5) = 88000 (1.01625)^{20} \] \[ A(5) \approx 121,477 \] 4. For \( t = 10 \): \[ A(10) = 88000 (1.01625)^{40} \] \[ A(10) \approx 167,792 \] ### d) When will the amount of money in the account reach $300,000? We need to solve for \( t \) in the equation \( A(t) = 300,000 \): \[ 300000 = 88000 (1.01625)^{4t} \] \[ \frac{300000}{88000} = (1.01625)^{4t} \] \[ 3.4091 \approx (1.01625)^{4t} \] Taking the natural logarithm of both sides: \[ \ln(3.4091) = 4t \cdot \ln(1.01625) \] \[ t = \frac{\ln(3.4091)}{4 \cdot \ln(1.01625)} \] \[ t \approx 17.33 \] In approximately 17.33 years, the amount will reach $300,000.