excans:61ffade529: Difference between revisions

Latest revision as of 01:34, 18 January 2024

Answer: C

Let [math]l_{80}=1000 \Rightarrow l_{81}=960 \Rightarrow l_{82}=902.4[/math]

[[math]]\textrm{Answer} \, =\frac{l_{81.1}-l_{81.6}}{l_{80.6}}=\frac{(960)^{0.9}(902.4)^{0.1}-(960)^{0.4}(902.4)^{0.6}}{(0.4)(1000)+(0.6)(960)}=0.02978[[/math]]

The value for [math]l_{80}[/math] is arbitrary. Any other starting value gives the same result.

Another form for the survivors between 81 and 82 would be

[math]\mu=-\ln (902 / 960)=0.0095833[/math];

[math]l_{81.1}=960 \times e^{-0.1 \times 0.0095833}=959.08 ;[/math]

[math]l_{81.6}=960 \times e^{-0.6 \times 0.0095833}=954.496[/math].

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-'''Answer: A'''
+'''Answer: C'''
-<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>
+Let <math>l_{80}=1000 \Rightarrow l_{81}=960 \Rightarrow l_{82}=902.4</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>
+<math display = "block">\textrm{Answer} \, =\frac{l_{81.1}-l_{81.6}}{l_{80.6}}=\frac{(960)^{0.9}(902.4)^{0.1}-(960)^{0.4}(902.4)^{0.6}}{(0.4)(1000)+(0.6)(960)}=0.02978</math>
+The value for <math>l_{80}</math> is arbitrary. Any other starting value gives the same result.
+Another form for the survivors between 81 and 82 would be
+<math>\mu=-\ln (902 / 960)=0.0095833</math>;
+<math>l_{81.1}=960 \times e^{-0.1 \times 0.0095833}=959.08 ;</math>
+<math>l_{81.6}=960 \times e^{-0.6 \times 0.0095833}=954.496</math>.
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