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Find the final amount of money in an account if \( \$ 1,200 \) is deposited at \( 2.5 \% \) interest compounded semi-annually and the money is left for 9 years. The final amount is \( \$ \square \) Round to the nearest cent. Question Help: \( \square \) Message instructor Calculator Submit Question

Ask by Mcfarlane Tyler. in the United States
Feb 27,2025

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Answer

The final amount in the account is approximately \$1,500.47.

Solution

To find the final amount A in an account where a principal P is compounded, we use the compound interest formula:   A = P(1 + r/n)^(n*t) Where:  P = initial deposit (in dollars)  r = annual interest rate (in decimal)  n = number of compounding periods per year  t = number of years For this problem:  P = 1200 dollars  r = 2.5% = 0.025  n = 2 (since interest is compounded semi-annually)  t = 9 years Step 1: Substitute the values into the formula:  A = 1200 * (1 + 0.025/2)^(2*9) Step 2: Simplify the inside of the parentheses:  0.025/2 = 0.0125, so:  A = 1200 * (1.0125)^(18) Step 3: Calculate (1.0125)^(18)  Using a calculator, (1.0125)^(18) ≈ 1.25039 Step 4: Compute the final amount:  A ≈ 1200 * 1.25039 ≈ 1500.468 Step 5: Round to the nearest cent:  A ≈ $1,500.47 Thus, the final amount in the account after 9 years is approximately $1,500.47.

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To find the final amount, you can use the compound interest formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \(A\) is the amount of money accumulated after n years, including interest, - \(P\) is the principal amount (the initial amount of money), - \(r\) is the annual interest rate (decimal), - \(n\) is the number of times that interest is compounded per year, - \(t\) is the number of years the money is invested or borrowed. Plugging in the values: - \(P = 1200\) - \(r = 0.025\) (which is \(2.5\%\)), - \(n = 2\) (because the interest is compounded semi-annually), - \(t = 9\). Now, substituting those values into the formula: \[ A = 1200 \left(1 + \frac{0.025}{2}\right)^{2 \cdot 9} \] \[ A = 1200 \left(1 + 0.0125\right)^{18} \] \[ A = 1200 \left(1.0125\right)^{18} \] Calculating \( (1.0125)^{18} \) gives approximately \(1.2434\). Now, calculate \(A\): \[ A \approx 1200 \times 1.2434 \approx 1492.08 \] Thus, the final amount in the account after 9 years is approximately \( \$1492.08 \). So, rounding to the nearest cent, the final amount is \( \$ \mathbf{1492.08} \).

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