Hunter invested \( \$ 580 \) in an account paying an interest rate of \( 4.6 \% \) compounded monthly, Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 19 years?

To calculate the future value of Hunter's investment, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial 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. Plugging in the values: - \( P = 580 \) - \( r = 0.046 \) - \( n = 12 \) (monthly compounding) - \( t = 19 \) We get: \[ A = 580 \left(1 + \frac{0.046}{12}\right)^{12 \times 19} \] Calculating the inside of the parentheses: \[ A = 580 \left(1 + 0.00383333\right)^{228} \] \[ A = 580 \left(1.00383333\right)^{228} \] Now calculating \( (1.00383333)^{228} \): \[ (1.00383333)^{228} \approx 2.4846 \] Finally, substituting back into the formula: \[ A \approx 580 \times 2.4846 \] \[ A \approx 1440.06 \] Rounding to the nearest dollar, the amount in the account after 19 years would be approximately \( \$ 1440 \).