1. **Problem 1:** How much money should be invested to have 8000 in 3 years at 6% interest compounded quarterly? 2. The compound interest formula is $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after time $t$ - $P$ is the principal (initial investment) - $r$ is the annual interest rate (decimal) - $n$ is the number of compounding periods per year - $t$ is the time in years 3. We want to find $P$ given $A=8000$, $r=0.06$, $n=4$, $t=3$. 4. Rearranging for $P$: $$P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}$$ 5. Substitute values: $$P = \frac{8000}{\left(1 + \frac{0.06}{4}\right)^{4 \times 3}} = \frac{8000}{\left(1 + 0.015\right)^{12}} = \frac{8000}{(1.015)^{12}}$$ 6. Calculate $(1.015)^{12}$: $$ (1.015)^{12} \approx 1.195618 $$ 7. Calculate $P$: $$ P = \frac{8000}{1.195618} \approx 6691.10 $$ 8. **Answer for Problem 1:** $6691.10$ 9. **Problem 2:** What interest will be earned if 4800 is invested for 7 years at 12% compounded monthly? 10. Use the same formula with $P=4800$, $r=0.12$, $n=12$, $t=7$. 11. Calculate amount $A$: $$ A = 4800 \left(1 + \frac{0.12}{12}\right)^{12 \times 7} = 4800 (1.01)^{84} $$ 12. Calculate $(1.01)^{84}$: $$ (1.01)^{84} \approx 2.30144 $$ 13. Calculate $A$: $$ A = 4800 \times 2.30144 \approx 11047.00 $$ 14. Interest earned is $A - P$: $$ 11047.00 - 4800 = 6247.00 $$ 15. The closest answer given is $11072.27$ which likely includes rounding differences; the correct interest earned is approximately $11072.27 - 4800 = 6272.27$ (recalculate precisely below). 16. Recalculate precisely: $$ (1.01)^{84} = e^{84 \ln(1.01)} \approx e^{0.835} \approx 2.30506 $$ $$ A = 4800 \times 2.30506 = 11064.29 $$ $$ \text{Interest} = 11064.29 - 4800 = 6264.29 $$ 17. The provided answer $11072.27$ is the total amount, so interest earned is $11072.27 - 4800 = 6272.27$. 18. **Problem 3:** How much money should be invested to have 5000 in 5 years at 9% interest compounded quarterly? 19. Use formula with $A=5000$, $r=0.09$, $n=4$, $t=5$. 20. Calculate $P$: $$ P = \frac{5000}{\left(1 + \frac{0.09}{4}\right)^{4 \times 5}} = \frac{5000}{(1.0225)^{20}} $$ 21. Calculate $(1.0225)^{20}$: $$ (1.0225)^{20} \approx 1.551328 $$ 22. Calculate $P$: $$ P = \frac{5000}{1.551328} \approx 3204.08 $$ 23. **Answer for Problem 3:** $3204.08$