excans:B8d33a4aef: Difference between revisions
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(Created page with "'''Answer: B''' Under constant force over each year of age, <math>l_{x+k}=\left(l_{x}\right)^{1-k}\left(l_{x+1}\right)^{k}</math> for <math>x</math> an integer and <math>0 \leq k \leq 1</math>. <math display="block"> \begin{aligned} & { }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}} \\ & l_{[60]+0.75}=(80,000)^{0.25}(79,000)^{0.75}=79,249 \\ & l_{[60]+2.75}=(77,000)^{0.25}(74,000)^{0.75}=74,739 \\ & l_{[60]+5.75}=(67,000)^{0.25}(65,000)^{...") |
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<math>{ }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}}=\frac{74,739-65,494}{79,249}=0.11679</math> | <math display = "block">{ }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}}=\frac{74,739-65,494}{79,249}=0.11679</math> | ||
<math>1000_{2 \mid 3} q_{[60]+0.75}=116.8</math> | <math display = "block">1000_{2 \mid 3} q_{[60]+0.75}=116.8</math> | ||
Revision as of 00:01, 16 January 2024
Answer: B
Under constant force over each year of age, [math]l_{x+k}=\left(l_{x}\right)^{1-k}\left(l_{x+1}\right)^{k}[/math] for [math]x[/math] an integer and [math]0 \leq k \leq 1[/math].
[[math]]
\begin{aligned}
& { }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}} \\
& l_{[60]+0.75}=(80,000)^{0.25}(79,000)^{0.75}=79,249 \\
& l_{[60]+2.75}=(77,000)^{0.25}(74,000)^{0.75}=74,739 \\
& l_{[60]+5.75}=(67,000)^{0.25}(65,000)^{0.75}=65,494
\end{aligned}
[[/math]]
[[math]]{ }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}}=\frac{74,739-65,494}{79,249}=0.11679[[/math]]
[[math]]1000_{2 \mid 3} q_{[60]+0.75}=116.8[[/math]]