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(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>.") |
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Therefore, <math>1000 \mu_{35}=(1000) \frac{1}{3(25)}=\frac{1000}{75}=13.3</math>. | Therefore, <math>1000 \mu_{35}=(1000) \frac{1}{3(25)}=\frac{1000}{75}=13.3</math>. | ||
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Latest revision as of 01:34, 18 January 2024
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].
Copyright 2024 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.