1. **State the problem:** Compare the final amounts of investing 1250 at 8% compounded monthly for 11 years and at 14% compounded monthly for 11 years. 2. **Formula used:** The compound interest formula is $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount of money accumulated after $t$ years, including interest. - $P$ is the principal amount (initial investment). - $r$ is the annual interest rate (decimal). - $n$ is the number of times interest is compounded per year. - $t$ is the time the money is invested for in years. 3. **Calculate for 8% interest:** - $P = 1250$ - $r = 0.08$ - $n = 12$ - $t = 11$ $$A = 1250\left(1 + \frac{0.08}{12}\right)^{12 \times 11} = 1250\left(1 + 0.0066667\right)^{132}$$ 4. **Calculate intermediate value:** $$1 + 0.0066667 = 1.0066667$$ 5. **Calculate power:** $$1.0066667^{132} \approx 2.4039$$ 6. **Calculate final amount:** $$A = 1250 \times 2.4039 = 3004.84$$ 7. **Calculate for 14% interest:** - $r = 0.14$ $$A = 1250\left(1 + \frac{0.14}{12}\right)^{12 \times 11} = 1250\left(1 + 0.0116667\right)^{132}$$ 8. **Calculate intermediate value:** $$1 + 0.0116667 = 1.0116667$$ 9. **Calculate power:** $$1.0116667^{132} \approx 4.6232$$ 10. **Calculate final amount:** $$A = 1250 \times 4.6232 = 5778.99$$ 11. **Calculate difference:** $$5778.99 - 3004.84 = 2774.15$$ **Note:** The user stated the difference as 5949.32, but the correct difference based on the calculations is 2774.15. **Final answers:** - Amount at 8%: $3004.84$ - Amount at 14%: $5778.99$ - Difference: $2774.15$