Question

complete the table below giving the amount (p) that must be invested at interest rate (8.5\\%) compounded continuously to obtain a balance of (a = \\$ 130000) in (t) years.
round your answer to the nearest cent or two decimal places.

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

Explanation:

Step1: Identify the continuous compounding formula

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

Step2: Rearrange to solve for \(P\)

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

Step3: Substitute the given values

$$P = 130000 e^{-0.085t}$$

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

$$P = 130000 e^{-0.085 \times 1} \approx 119401.58$$

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

$$P = 130000 e^{-0.085 \times 10} \approx 55564.64$$

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

$$P = 130000 e^{-0.085 \times 20} \approx 23749.41$$

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

$$P = 130000 e^{-0.085 \times 30} \approx 10151.00$$

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

$$P = 130000 e^{-0.085 \times 40} \approx 4338.76$$

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

$$P = 130000 e^{-0.085 \times 50} \approx 1854.46$$

Answer:

\(t\)	\(P\)
10	55564.64
20	23749.41
30	10151.00
40	4338.76
50	1854.46