excans:Bd3576e403: Difference between revisions

Latest revision as of 02:34, 18 January 2024

Answer: A

[math]f_{x}(t)=-\frac{d}{d t} S_{x}(t)=-\frac{d}{d t}\left(e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)}\right)[/math]

[math]=-e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)} \cdot\left(-\frac{B}{\ln c} \cdot c^{x}\right) \cdot c^{t} \cdot \ln c[/math]

[math]=e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)} \cdot B c^{x+t}[/math]

[math]=0.00027 \times 1.1^{x+t} \cdot e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{x}\right)\left(1.1^{t}-1\right)}[/math]

[math]f_{50}(10)=0.00027 \times 1.1^{50+10} \cdot e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{50}\right)\left(1.1^{10}-1\right)}=0.04839[/math]

Alternative Solution:

[math]f_{x}(t)={ }_{t} p_{x} \cdot \mu_{x+t}[/math]

Then we can use the formulas given for Makeham with [math]A=0, B=0.00027[/math] and [math]c=1.1[/math]

[math]f_{x}(t)=\left(e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{50}\right)\left(1.1^{10}-1\right)}\right)\left(0.00027 \times 1.1^{50+10}\right)=0.04839[/math]

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 <math>f_{x}(t)=\left(e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{50}\right)\left(1.1^{10}-1\right)}\right)\left(0.00027 \times 1.1^{50+10}\right)=0.04839</math>
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