consider the following sequence of year - end cash flows:\neoy 1 2 3 4 5\ncash flow $6,000 $14,000 $22,000…

Explanation:

Step1: Recall present - worth formula

The present - worth formula for a single cash - flow $F$ at the end of year $n$ with an interest rate $i$ is $P = \frac{F}{(1 + i)^n}$.

Step2: Calculate present worth of each cash - flow

For $n = 1$, $F_1=6000$, $P_1=\frac{6000}{(1 + 0.08)^1}=\frac{6000}{1.08}\approx5555.56$. For $n = 2$, $F_2 = 14000$, $P_2=\frac{14000}{(1 + 0.08)^2}=\frac{14000}{1.1664}\approx11994.17$. For $n = 3$, $F_3 = 22000$, $P_3=\frac{22000}{(1 + 0.08)^3}=\frac{22000}{1.259712}\approx17464.39$. For $n = 4$, $F_4 = 30000$, $P_4=\frac{30000}{(1 + 0.08)^4}=\frac{30000}{1.36048896}\approx22050.42$. For $n = 5$, $F_5 = 38000$, $P_5=\frac{38000}{(1 + 0.08)^5}=\frac{38000}{1.4693280768}\approx25853.46$.

Step3: Sum up present worths

$P=P_1 + P_2+P_3+P_4+P_5$ $P\approx5555.56+11994.17+17464.39+22050.42+25853.46 = 82918.0$. Rounding differences may occur. The closest value is $$82,936$.

Answer:

E. $82,936$