1. **State the problem:** Enola and Alexandra each invest 67000 in different accounts with different interest rates and compounding methods. We want to find how much more money Enola has than Alexandra after 10 years. 2. **Formulas used:** - For continuous compounding: $$A = P e^{rt}$$ where $P$ is principal, $r$ is annual interest rate (decimal), $t$ is time in years. - For monthly compounding: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where $n$ is number of compounding periods per year. 3. **Convert interest rates to decimals:** - Enola's rate: 6 1/4% = 6.25% = 0.0625 - Alexandra's rate: 5 7/8% = 5.875% = 0.05875 4. **Calculate Enola's amount after 10 years:** $$A_E = 67000 \times e^{0.0625 \times 10} = 67000 \times e^{0.625}$$ Calculate $e^{0.625} \approx 1.86825$ $$A_E \approx 67000 \times 1.86825 = 125140.75$$ 5. **Calculate Alexandra's amount after 10 years:** Monthly compounding means $n=12$. $$A_A = 67000 \times \left(1 + \frac{0.05875}{12}\right)^{12 \times 10} = 67000 \times \left(1 + 0.0048958\right)^{120}$$ Calculate base: $1.0048958$ Calculate exponent: $1.0048958^{120} \approx 1.8194$ $$A_A \approx 67000 \times 1.8194 = 121697.8$$ 6. **Find the difference:** $$\text{Difference} = A_E - A_A = 125140.75 - 121697.8 = 3443$$ 7. **Final answer:** Enola would have approximately 3443 more than Alexandra after 10 years.