Revision as of 22:09, 19 January 2024 by Admin (Created page with "'''Answer: D''' Let <math>P</math> be the premium per 1 of insurance. <math display="block"> \begin{aligned} & P \ddot{a}_{50: \overline{10}}=P(I A)_{50: 10}^{1}+{ }_{10} E_{50} A_{60} \\ & \ddot{a}_{50: 10 \mid}=\ddot{a}_{50}-{ }_{10} E_{50} \ddot{a}_{60}=17.0-0.60 \times 15.0=8 \\ & A_{60}=1-d \ddot{a}_{60}=1-\left(\frac{0.05}{1.05}\right) 15=0.285714 \\ & P\left(\ddot{a}_{50: \overline{10}}-(I A)_{50: 10}^{1}\right)={ }_{10} E_{50} A_{60} \\ & P=\frac{10}{\ddot{a}_...")
Exercise
ABy Admin
Jan 19'24
Answer
Answer: D
Let [math]P[/math] be the premium per 1 of insurance.
[[math]]
\begin{aligned}
& P \ddot{a}_{50: \overline{10}}=P(I A)_{50: 10}^{1}+{ }_{10} E_{50} A_{60} \\
& \ddot{a}_{50: 10 \mid}=\ddot{a}_{50}-{ }_{10} E_{50} \ddot{a}_{60}=17.0-0.60 \times 15.0=8 \\
& A_{60}=1-d \ddot{a}_{60}=1-\left(\frac{0.05}{1.05}\right) 15=0.285714 \\
& P\left(\ddot{a}_{50: \overline{10}}-(I A)_{50: 10}^{1}\right)={ }_{10} E_{50} A_{60} \\
& P=\frac{10}{\ddot{a}_{50: \overline{10}}-(I A)_{50: 10}^{1}}=\frac{0.6 \times 0.285714}{8-0.15}=0.021838 \\
& 100 P=2.18
\end{aligned}
[[/math]]