Question

complete the table below giving the amount (p) that must be invested at interest rate (10.5%) compounded continuously to obtain a balance of (a = \\$ 200000) in (t) years.
round your answer to the nearest cent or two decimal places.

\

$$\begin{tabular}{|c|c|} \\hline (t) & (p) \\\\ \\hline 1 & \\\\ \\hline 10 & \\\\ \\hline 20 & \\\\ \\hline 30 & \\\\ \\hline 40 & \\\\ \\hline 50 & \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Identify the continuous compounding formula

$$A = P e^{rt}$$

Step2: Rearrange the formula to solve for \(P\)

$$P = A e^{-rt}$$

Step3: Substitute the given values into the formula

$$P = 200000 e^{-0.105t}$$

Step4: Calculate \(P\) for \(t = 1\)

$$P = 200000 e^{-0.105 \times 1} \approx 180060.05$$

Step5: Calculate \(P\) for \(t = 10\)

$$P = 200000 e^{-0.105 \times 10} \approx 70000.52$$

Step6: Calculate \(P\) for \(t = 20\)

$$P = 200000 e^{-0.105 \times 20} \approx 24500.36$$

Step7: Calculate \(P\) for \(t = 30\)

$$P = 200000 e^{-0.105 \times 30} \approx 8575.25$$

Step8: Calculate \(P\) for \(t = 40\)

$$P = 200000 e^{-0.105 \times 40} \approx 3001.38$$

Step9: Calculate \(P\) for \(t = 50\)

$$P = 200000 e^{-0.105 \times 50} \approx 1050.50$$

Answer:

\(t\)	\(P\)
10	\$70,000.52
20	\$24,500.36
30	\$8,575.25
40	\$3,001.38
50	\$1,050.50