Cooper invested 1,100 in an account paying an interest rate of 3.8% compounded daily. Assuming no deposits or withdrawals are made, how much money, to the nearest cent, would be in the account after 10 years? Answer Attemptiout of 2

To solve this problem, we need to use the formula for compound interest:<br /><br />\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]<br /><br />Where:<br />- \( A \) is the amount of money in the account after \( t \) years.<br />- \( P \) is the principal amount (initial investment), which is $1,100.<br />- \( r \) is the annual interest rate (as a decimal), which is 0.038.<br />- \( n \) is the number of times the interest is compounded per year, which is 365 (daily compounding).<br />- \( t \) is the number of years, which is 10.<br /><br />Plugging in the values, we get:<br /><br />\[ A = 1100 \left(1 + \frac{0.038}{365}\right)^{365 \times 10} \]<br /><br />First, calculate the daily interest rate:<br /><br />\[ \frac{0.038}{365} \approx 0.00010411 \]<br /><br />Next, calculate the exponent:<br /><br />\[ 365 \times 10 = 3650 \]<br /><br />Now, calculate the base raised to the power of 3650:<br /><br />\[ \left(1 + 0.00010411\right)^{3650} \approx 1.4816 \]<br /><br />Finally, multiply this by the principal amount:<br /><br />\[ A = 1100 \times 1.4816 \approx 1627.76 \]<br /><br />So, the amount of money in the account after 10 years, to the nearest cent, is approximately $1627.76.