1. **State the problem:** Bob invests 1100 dollars in a 3-year CD with an annual interest rate of 1.28%, compounded daily. We need to find: (a) The amount of money in Bob's account after 3 years. (b) The interest earned after 3 years. 2. **Formula for compound interest with daily compounding:** $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where - $A$ is the amount after time $t$, - $P = 1100$ is the principal, - $r = 0.0128$ is the annual interest rate as a decimal, - $n = 365$ is the number of compounding periods per year (daily), - $t = 3$ years. 3. **Calculate the amount after 3 years:** $$A = 1100 \left(1 + \frac{0.0128}{365}\right)^{365 \times 3} = 1100 \left(1 + 0.000035068\right)^{1095}$$ Calculate the base inside the parenthesis: $$1 + 0.000035068 = 1.000035068$$ Now raise to the power of 1095: $$1.000035068^{1095} \approx e^{1095 \times \ln(1.000035068)}$$ Calculate the exponent: $$1095 \times \ln(1.000035068) \approx 1095 \times 0.000035068 = 0.03838$$ Thus: $$A \approx 1100 \times e^{0.03838} \approx 1100 \times 1.03915 = 1143.06$$ 4. **Calculate the interest earned:** $$\text{Interest} = A - P = 1143.06 - 1100 = 43.06$$ **Final answers:** (a) Amount after 3 years: $1143.06$ (b) Interest earned after 3 years: $43.06$