1. **Problem Statement:** Calculate the Present Value (P) for each row given the future value (Ra), interest rate (r), compounding frequency (m), and time (t). 2. **Formula:** The Present Value formula for compound interest is: $$P = \frac{Ra}{\left(1 + \frac{r}{m}\right)^{m \times t}}$$ where: - $Ra$ is the future value - $r$ is the annual interest rate (decimal) - $m$ is the number of compounding periods per year - $t$ is the number of years 3. **Important Notes:** - Convert percentage rates to decimals by dividing by 100. - Use the correct $m$ based on compounding frequency: - annually: $m=1$ - semi-annually: $m=2$ - quarterly: $m=4$ - monthly: $m=12$ - Round answers to the nearest hundredths. 4. **Calculations:** **Row 2:** - $Ra=1100$, $r=0.08$, $m=1$, $t=20$ - $$P=\frac{1100}{(1+\frac{0.08}{1})^{1\times20}}=\frac{1100}{(1.08)^{20}}$$ - Calculate $(1.08)^{20} \approx 4.66096$ - $$P=\frac{1100}{4.66096} \approx 236.00$$ **Row 3:** - $Ra=2200$, $r=0.09$, $m=2$, $t=15$ - $$P=\frac{2200}{(1+\frac{0.09}{2})^{2\times15}}=\frac{2200}{(1.045)^{30}}$$ - Calculate $(1.045)^{30} \approx 3.80611$ - $$P=\frac{2200}{3.80611} \approx 577.96$$ **Row 4:** - $Ra=3300$, $r=0.10$, $m=4$, $t=10$ - $$P=\frac{3300}{(1+\frac{0.10}{4})^{4\times10}}=\frac{3300}{(1.025)^{40}}$$ - Calculate $(1.025)^{40} \approx 2.68506$ - $$P=\frac{3300}{2.68506} \approx 1229.44$$ **Row 5:** - $Ra=4400$, $r=0.11$, $m=12$, $t=5$ - $$P=\frac{4400}{(1+\frac{0.11}{12})^{12\times5}}=\frac{4400}{(1.0091667)^{60}}$$ - Calculate $(1.0091667)^{60} \approx 1.74742$ - $$P=\frac{4400}{1.74742} \approx 2517.15$$ **Row 6:** - $Ra=5500$, $r=0.12$, $m=1$, $t=5$ - $$P=\frac{5500}{(1+\frac{0.12}{1})^{1\times5}}=\frac{5500}{(1.12)^{5}}$$ - Calculate $(1.12)^{5} \approx 1.76234$ - $$P=\frac{5500}{1.76234} \approx 3121.44$$ 5. **Final Table Present Values:** - Row 2: 236.00 - Row 3: 577.96 - Row 4: 1229.44 - Row 5: 2517.15 - Row 6: 3121.44 These values are rounded to the nearest hundredths as requested.