Revision as of 20:28, 15 January 2024 by Admin (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>")
Exercise
ABy Admin
Jan 15'24
Answer
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]