Question 1 of 15, Step 1 of 1 Angela deposits \( \$ 4000 \) into an account with an APR of \( 4.6 \% \) for 12 years. Find the future value of the account if interest is compounded monthly. Round your answer to the nearest hundredth, if necessary.

To find the future value of an account with compound interest, we can use the formula: \[ FV = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( FV \) is the future value of the investment/loan, including interest, - \( P \) is the principal investment amount (the initial deposit or loan amount), - \( r \) is the annual interest rate (decimal), - \( n \) is the number of times that interest is compounded per unit \( t \), - \( t \) is the time the money is invested or borrowed for, in years. Given: - \( P = 4000 \) - \( r = 4.6\% = 0.046 \) - \( n = 12 \) (since interest is compounded monthly) - \( t = 12 \) Now, we can substitute these values into the formula to calculate the future value \( FV \): \[ FV = 4000 \left(1 + \frac{0.046}{12}\right)^{12 \times 12} \] Let's calculate this step by step. 1. Calculate \( \frac{r}{n} \): \[ \frac{0.046}{12} \] 2. Calculate \( nt \): \[ 12 \times 12 \] 3. Substitute these values into the formula and calculate \( FV \). Now, I will perform these calculations. Calculate the value by following steps: - step0: Calculate: \(12\times 12\) - step1: Multiply the numbers: \(144\) Calculate or simplify the expression \( 0.046/12 \). Calculate the value by following steps: - step0: Calculate: \(\frac{0.046}{12}\) - step1: Convert the expressions: \(\frac{\frac{23}{500}}{12}\) - step2: Multiply by the reciprocal: \(\frac{23}{500}\times \frac{1}{12}\) - step3: Multiply the fractions: \(\frac{23}{500\times 12}\) - step4: Multiply: \(\frac{23}{6000}\) Calculate or simplify the expression \( 4000*(1+0.00383333)^144 \). Calculate the value by following steps: - step0: Calculate: \(4000\left(1+0.00383333\right)^{144}\) - step1: Add the numbers: \(4000\times 1.00383333^{144}\) - step2: Convert the expressions: \(4000\left(\frac{100383333}{100000000}\right)^{144}\) - step3: Simplify: \(4000\times \frac{100383333^{144}}{100000000^{144}}\) - step4: Rewrite the expression: \(4000\times \frac{100383333^{144}}{4000^{144}\times 25000^{144}}\) - step5: Reduce the numbers: \(1\times \frac{100383333^{144}}{4000^{143}\times 25000^{144}}\) - step6: Multiply the fractions: \(\frac{100383333^{144}}{4000^{143}\times 25000^{144}}\) Now that we have the necessary calculations: 1. The monthly interest rate \( \frac{r}{n} \) is approximately \( 0.00383333 \). 2. The total number of compounding periods \( nt \) is \( 144 \). Now we can substitute these values into the future value formula: \[ FV = 4000 \left(1 + 0.00383333\right)^{144} \] Calculating this gives us: \[ FV \approx 6939.56 \] Thus, the future value of the account after 12 years, with interest compounded monthly, is approximately **\$6939.56**.