excans:Dcb1e2fa8b: Difference between revisions
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(Created page with "'''Answer: C''' <math display="block"> \begin{aligned} & { }_{3.5} p_{[61]}-{ }_{3.5} p_{[60]+1}={ }_{0.5} p_{64}\left({ }_{3} p_{[61]}-{ }_{3} p_{[60]+1}\right) \\ & =\left(\frac{l_{65}}{l_{64}}\right)^{0.5}\left(\frac{l_{64}}{l_{[61]}}-\frac{l_{64}}{l_{[60]+1}}\right) \\ & =\left(\frac{4016}{5737}\right)^{0.5}\left(\frac{5737}{8654}-\frac{5737}{9600}\right) \\ & =0.05466 \end{aligned} </math>") |
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Revision as of 02:33, 18 January 2024
Answer: C
[[math]]
\begin{aligned}
& { }_{3.5} p_{[61]}-{ }_{3.5} p_{[60]+1}={ }_{0.5} p_{64}\left({ }_{3} p_{[61]}-{ }_{3} p_{[60]+1}\right) \\
& =\left(\frac{l_{65}}{l_{64}}\right)^{0.5}\left(\frac{l_{64}}{l_{[61]}}-\frac{l_{64}}{l_{[60]+1}}\right) \\
& =\left(\frac{4016}{5737}\right)^{0.5}\left(\frac{5737}{8654}-\frac{5737}{9600}\right) \\
& =0.05466
\end{aligned}
[[/math]]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.