Exercise
ABy Admin
Jan 19'24
Answer
Answer: B
By the equivalence principle,
[[math]]
4500 \bar{a}_{x: \overline{20}}=100,000 \bar{A}_{x: \overline{20}}^{1}+R \bar{a}_{x: \overline{20}}
[[/math]]
where
[[math]]
\begin{aligned}
& \bar{A}_{x: 20 \mid}^{1}=\frac{\mu}{\mu+\delta}\left(1-e^{-20(\mu+\delta)}\right)=\frac{0.04}{0.12}\left(1-e^{-20(0.12)}\right)=0.3031 \\
& \bar{a}_{x: 20 \mid}=\frac{1-e^{-20(\mu+\delta)}}{\mu+\delta}=\frac{1-e^{-20(0.12)}}{0.12}=7.5774
\end{aligned}
[[/math]]
Solving for [math]R[/math], we have
[[math]]
R=4500-100,000\left(\frac{0.3031}{7.5774}\right)=500
[[/math]]