1. **Problem:** Ferb invests 400 at 3.00% interest compounded quarterly for 3 years. Find the final amount. 2. **Formula:** Compound interest formula is $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where $P$ is principal, $r$ is annual rate (decimal), $n$ is compounding periods per year, and $t$ is years. 3. **Substitute values:** $P=400$, $r=0.03$, $n=4$, $t=3$. 4. Calculate inside the parentheses: $$1 + \frac{0.03}{4} = 1 + 0.0075 = 1.0075$$ 5. Calculate exponent: $$nt = 4 \times 3 = 12$$ 6. Calculate amount: $$A = 400 \times 1.0075^{12}$$ 7. Calculate power: $$1.0075^{12} \approx 1.093443$$ 8. Multiply: $$400 \times 1.093443 = 437.3772$$ 9. Round to nearest cent: $$437.38$$ --- 1. **Problem:** $210 was invested at simple interest for 3 years and returned $277. Find the interest rate as a percent. 2. **Formula:** Simple interest formula is $$A = P(1 + rt)$$ where $A$ is amount, $P$ is principal, $r$ is rate (decimal), $t$ is time in years. 3. Substitute values: $$277 = 210(1 + 3r)$$ 4. Divide both sides by 210: $$\frac{277}{210} = 1 + 3r$$ 5. Simplify fraction: $$\cancel{\frac{277}{210}} = 1.3190$$ 6. Subtract 1: $$1.3190 - 1 = 3r$$ 7. Simplify: $$0.3190 = 3r$$ 8. Divide both sides by 3: $$r = \frac{0.3190}{3} = 0.1063$$ 9. Convert to percent: $$0.1063 \times 100 = 10.63$$ 10. Round to nearest tenth: $$10.6$$ **Final answers:** - Compound interest amount: 437.38 - Simple interest rate: 10.6