Revision as of 19:34, 15 January 2024 by Admin (Created page with "'''Answer: C''' <math display="block"> \begin{aligned} \mu_{x} & =-\frac{d}{d_{x}} \ln S_{0}(x)=-\frac{1}{3} \frac{d}{d_{x}} \ln \left(1-\frac{x}{60}\right) \\ & =\frac{1}{180}\left(1-\frac{x}{60}\right)^{-1}=\frac{1}{3(60-x)} \end{aligned} </math> Therefore, <math>1000 \mu_{35}=(1000) \frac{1}{3(25)}=\frac{1000}{75}=13.3</math>.")
Exercise
ABy Admin
Jan 15'24
Answer
Answer: C
[[math]]
\begin{aligned}
\mu_{x} & =-\frac{d}{d_{x}} \ln S_{0}(x)=-\frac{1}{3} \frac{d}{d_{x}} \ln \left(1-\frac{x}{60}\right) \\
& =\frac{1}{180}\left(1-\frac{x}{60}\right)^{-1}=\frac{1}{3(60-x)}
\end{aligned}
[[/math]]
Therefore, [math]1000 \mu_{35}=(1000) \frac{1}{3(25)}=\frac{1000}{75}=13.3[/math].