1. **State the problem:** Bob invests 1100 for 3 years at an annual interest rate of 1.28% compounded daily. We need to find: (a) The future value in Bob's account after 3 years. (b) The amount of interest earned after 3 years. 2. **Identify the formula:** For compound interest compounded daily, the formula is: $$ A = P\left(1 + \frac{r}{n}\right)^{nt} $$ where: - $A$ is the amount after time $t$ - $P = 1100$ is the principal - $r = 0.0128$ (1.28% as a decimal) is the annual interest rate - $n = 365$ is the number of compounding periods per year (daily) - $t = 3$ years is the time 3. **Calculate the amount $A$: ** Plug values into the formula: $$ A = 1100 \left(1 + \frac{0.0128}{365}\right)^{365 \times 3} $$ Calculate inside the parenthesis: $$ 1 + \frac{0.0128}{365} = 1 + 0.00003506849315 = 1.00003506849315 $$ Calculate the exponent: $$ 365 \times 3 = 1095 $$ Now compute: $$ A = 1100 \times (1.00003506849315)^{1095} $$ Calculate the power: $$ (1.00003506849315)^{1095} \approx e^{1095 \times 0.00003506849315} = e^{0.038385} \approx 1.039153 $$ So: $$ A \approx 1100 \times 1.039153 = 1143.0683 $$ Round to nearest cent: $$ A \approx 1143.07 $$ 4. **Calculate interest earned:** Interest $I = A - P$ $$ I = 1143.07 - 1100 = 43.07 $$ Final answers: - (a) Future value is $1143.07$ - (b) Interest earned is $43.07$