If 6600 dollars is invested at an interest rate of 5 percent per year, find the value of the investment at the end of 5 years for the following compounding methods. Round value to the nearest cent. (a) Annual: \( \$ \) (b) Semiannual: \( \$ \) (c) Monthly: \( \$ \) (d) Daily: \( \$ \)
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To calculate the future value of an investment using different compounding methods, we can use the 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 time the money is invested for in years. Given: - \( P = 6600 \) - \( r = 0.05 \) - \( t = 5 \) (a) For annual compounding (\( n = 1 \)): \[ A = 6600 \left(1 + \frac{0.05}{1}\right)^{1 \times 5} = 6600 \left(1 + 0.05\right)^{5} = 6600 \times (1.2762815625) \approx 8,067.86 \] (b) For semiannual compounding (\( n = 2 \)): \[ A = 6600 \left(1 + \frac{0.05}{2}\right)^{2 \times 5} = 6600 \left(1 + 0.025\right)^{10} = 6600 \times (1.280084528) \approx 8,129.56 \] (c) For monthly compounding (\( n = 12 \)): \[ A = 6600 \left(1 + \frac{0.05}{12}\right)^{12 \times 5} = 6600 \left(1 + 0.00416667\right)^{60} = 6600 \times (1.283682575) \approx 8,164.00 \] (d) For daily compounding (\( n = 365 \)): \[ A = 6600 \left(1 + \frac{0.05}{365}\right)^{365 \times 5} = 6600 \left(1 + 0.0001369863\right)^{1825} = 6600 \times (1.2840254167) \approx 8,195.96 \] So, rounding values to the nearest cent, we have: (a) \( 8,067.86 \) dollars (b) \( 8,129.56 \) dollars (c) \( 8,164.00 \) dollars (d) \( 8,195.96 \) dollars
