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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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Latest revision as of 02:34, 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.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.