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(Created page with "'''Answer: A''' <math>{ }_{10} p_{x}=\frac{l_{x+10}}{l_{x}}=e^{-\int_{0}^{10} \mu_{x+1} \cdot d t}=>\frac{400}{1000}=e^{-\int_{0}^{10} \beta t^{2} \cdot d t}=>0.4=e^{-\beta t^{3} / 3 b^{10}}</math> <math>==>0.4=e^{-\beta \cdot 100^{3} / 3}==>\ln (0.4)=-\beta\left(\frac{1000}{3}\right)==>\beta=-\ln (0.4)(.003)=0.0027489</math>") |
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<math>==>0.4=e^{-\beta \cdot 100^{3} / 3}==>\ln (0.4)=-\beta\left(\frac{1000}{3}\right)==>\beta=-\ln (0.4)(.003)=0.0027489</math> | <math>==>0.4=e^{-\beta \cdot 100^{3} / 3}==>\ln (0.4)=-\beta\left(\frac{1000}{3}\right)==>\beta=-\ln (0.4)(.003)=0.0027489</math> | ||
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Revision as of 02:33, 18 January 2024
Answer: A
[math]{ }_{10} p_{x}=\frac{l_{x+10}}{l_{x}}=e^{-\int_{0}^{10} \mu_{x+1} \cdot d t}=\gt\frac{400}{1000}=e^{-\int_{0}^{10} \beta t^{2} \cdot d t}=\gt0.4=e^{-\beta t^{3} / 3 b^{10}}[/math]
[math]==\gt0.4=e^{-\beta \cdot 100^{3} / 3}==\gt\ln (0.4)=-\beta\left(\frac{1000}{3}\right)==\gt\beta=-\ln (0.4)(.003)=0.0027489[/math]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.