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| rev | Admin | (Created page with "'''Answer: B''' Let <math>P</math> be the annual net premium <math>P=\frac{1000 \bar{A}_{x: n}}{\ddot{a}_{x: n}}=\frac{1000(0.192)}{\ddot{a}_{x: n}}</math> where <math>\ddot{a}_{x: n]}=\frac{1-A_{x: n}}{d}=\frac{(1.05)}{(0.05)}\left(1-A_{x: n]}^{1}-A_{x: n]}^{1}\right)</math> <math>A_{x: n}=\frac{i}{\delta}\left(A_{x: n}^{1}\right)+{ }_{n} E_{x}</math> <math>\Rightarrow 0.192=\frac{0.05}{0.04879}\left(A_{x: n}^{1}\right)+0.172</math> <math>\Rightarrow A_{x: n}^{1}...") | Jan 19'24 at 21:40 | +673 |