1. **Problem statement:** Compare two investments with different annual interest rates compounded quarterly to determine which yields more interest. 2. **Formula used:** The compound interest formula is $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after time $t$ years, - $P$ is the principal (initial amount), - $r$ is the annual interest rate (decimal), - $n$ is the number of compounding periods per year, - $t$ is the time in years. 3. **Given:** - Investment 1: $r_1 = 7.25\% = 0.0725$, $n=4$ (quarterly) - Investment 2: $r_2 = 6.75\% = 0.0675$, $n=4$ (quarterly) 4. **Assume:** $P=1$ (for comparison, principal cancels out), $t=1$ year. 5. **Calculate amount for Investment 1:** $$A_1 = 1 \times \left(1 + \frac{0.0725}{4}\right)^{4 \times 1} = \left(1 + 0.018125\right)^4 = 1.018125^4$$ Calculate: $$1.018125^4 = 1.0749$$ (approx) 6. **Calculate amount for Investment 2:** $$A_2 = 1 \times \left(1 + \frac{0.0675}{4}\right)^4 = \left(1 + 0.016875\right)^4 = 1.016875^4$$ Calculate: $$1.016875^4 = 1.0690$$ (approx) 7. **Compare:** $$A_1 = 1.0749 > A_2 = 1.0690$$ 8. **Conclusion:** The first investment with 7.25% compounded quarterly yields more interest after one year.