Jim invested \( \$ 5600 \) in an account that pays an annual interest rate of \( 2.2 \% \), compounded monthly. Answer e list of financial formulas. (a) Find the amount in the account after one year, assuming no withdrawals are made. Do not round any intermediate computations, and round your answer to the nearest cent. \( \$ \square \) (b) Find the effective annual interest rate, expressed as a percentage. Do not round any intermediate computations, and round your answer to the nearest hundredth of a percent. \( \square \% \)
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The Deep Dive
To find the amount in the account after one year with monthly compounding, we can use the formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount after time \( t \), - \( P \) is the principal amount (\$5600), - \( r \) is the annual interest rate (0.022), - \( n \) is the number of times interest is compounded per year (12 for monthly), - \( t \) is the time in years (1). Plugging in the values: \[ A = 5600 \left(1 + \frac{0.022}{12}\right)^{12 \times 1} \] \[ A = 5600 \left(1 + 0.00183333\right)^{12} \] \[ A = 5600 \left(1.00183333\right)^{12} \] \[ A \approx 5600 \times 1.02226434 \approx 5724.08 \] So after one year, the amount in the account is approximately \( \boxed{5724.08} \). For the effective annual interest rate (EAR), we can use the formula: \[ EAR = \left(1 + \frac{r}{n}\right)^{n} - 1 \] Substituting the values: \[ EAR = \left(1 + \frac{0.022}{12}\right)^{12} - 1 \] \[ EAR = \left(1.00183333\right)^{12} - 1 \] \[ EAR \approx 1.02226434 - 1 \approx 0.02226434 \] Converting this to a percentage: \[ EAR \approx 0.02226434 \times 100 \approx 2.23\% \] Thus, the effective annual interest rate is approximately \( \boxed{2.23\%} \).