Revision as of 23:18, 18 January 2024 by Admin (Created page with "'''Answer: B''' <math>A_{x: n \mid}^{1}={ }_{n} E_{x}</math> <math>A_{x}=A_{x: n}^{1}+{ }_{n} E_{x} A_{x+n}</math> <math>0.3=A_{x: n}^{1}+(0.35)(0.4) \Rightarrow A_{x: n}^{1}=0.16</math> <math>A_{x: n}=A_{x: n}^{1}+{ }_{n} E_{x}=0.16+0.35=0.51</math> <math>\ddot{a}_{x: n}=\frac{1-A_{x: n}}{d}=\frac{1-0.51}{(0.05 / 1.05)}=10.29</math> <math>a_{x: n}=\ddot{a}_{x: n}-1+{ }_{n} E_{x}=10.29-0.65=9.64</math> {{soacopyright|2024}}")
Exercise
Jan 18'24
Answer
Answer: B
[math]A_{x: n \mid}^{1}={ }_{n} E_{x}[/math]
[math]A_{x}=A_{x: n}^{1}+{ }_{n} E_{x} A_{x+n}[/math]
[math]0.3=A_{x: n}^{1}+(0.35)(0.4) \Rightarrow A_{x: n}^{1}=0.16[/math]
[math]A_{x: n}=A_{x: n}^{1}+{ }_{n} E_{x}=0.16+0.35=0.51[/math]
[math]\ddot{a}_{x: n}=\frac{1-A_{x: n}}{d}=\frac{1-0.51}{(0.05 / 1.05)}=10.29[/math]
[math]a_{x: n}=\ddot{a}_{x: n}-1+{ }_{n} E_{x}=10.29-0.65=9.64[/math]
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