Revision as of 23:30, 18 January 2024 by Admin (Created page with "'''Answer: A''' <math>\stackrel{\circ}{e}_{40}=\frac{1}{\mu}=50</math> So, receive <math>K</math> for 50 years guaranteed and for life thereafter. <math>10,000=K\left[\bar{a}_{\overline{50}}+{ }_{50 \mid} \bar{a}_{40}\right]</math> <math>\bar{a}_{\overline{50}}=\int_{0}^{50} e^{-\delta t}=\frac{1-e^{-50 \delta}}{\delta}=\frac{1-e^{-50(0.01)}}{0.01}=39.35</math> <math>{ }_{50} \bar{a}_{40}={ }_{50} E_{40} \bar{a}_{40+50}=e^{-(\delta+\mu) 50} \frac{1}{\mu+\delta}=e^{-1...")
Exercise
Jan 18'24
Answer
Answer: A
[math]\stackrel{\circ}{e}_{40}=\frac{1}{\mu}=50[/math] So, receive [math]K[/math] for 50 years guaranteed and for life thereafter.
[math]10,000=K\left[\bar{a}_{\overline{50}}+{ }_{50 \mid} \bar{a}_{40}\right][/math]
[math]\bar{a}_{\overline{50}}=\int_{0}^{50} e^{-\delta t}=\frac{1-e^{-50 \delta}}{\delta}=\frac{1-e^{-50(0.01)}}{0.01}=39.35[/math]
[math]{ }_{50} \bar{a}_{40}={ }_{50} E_{40} \bar{a}_{40+50}=e^{-(\delta+\mu) 50} \frac{1}{\mu+\delta}=e^{-1.5} \frac{1}{0.03}=7.44[/math]
[math]K=\frac{10,000}{39.35+7.44}=213.7[/math]
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