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Brenda invests in a savings account with a fixed annual interest rate of compounded times per year. What will the account balance be after years?
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Q. Brenda invests in a savings account with a fixed annual interest rate of compounded times per year. What will the account balance be after years?
- Identify Variables: First, we need to identify the variables for the compound interest formula, which is , where:A = the future value of the investment/loan, including interestP = the principal investment amount ($\$\(4\),\(848\))\(\newline\)r = the annual interest rate (decimal) (\(5\%\) or \(0.05\))\(\newline\)n = the number of times that interest is compounded per year (\(2\))\(\newline\)t = the time the money is invested for in years (\(6\))\(\newline\)Now we will plug these values into the formula to calculate the future value of Brenda's investment.
- Calculate Rate per Period: Calculate the rate per period by dividing the annual interest rate by the number of compounding periods per year. \(\frac{r}{n} = \frac{0.05}{2} = 0.025\)
- Calculate Total Compounding Periods: Calculate the total number of compounding periods by multiplying the number of years by the number of compounding periods per year. \(n_t = 6 \times 2 = 12\)
- Substitute Values into Formula: Substitute the values into the compound interest formula to calculate the future value.\(\newline\)\(A = 4848 \times (1 + 0.025)^{12}\)
- Calculate Value Inside Parentheses: Calculate the value inside the parentheses first. \(1 + 0.025 = 1.025\)
- Calculate Compound Factor: Now raise \(1.025\) to the power of \(12\) to find the compound factor.\(\newline\)\((1.025)^{12} \approx 1.34489\)
- Calculate Future Value: Multiply the principal amount by the compound factor to find the future value.\(\newline\)\(A = 4848 \times 1.34489 \approx 6519.97\)
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