Revision as of 23:31, 18 January 2024 by Admin (Created page with "'''Answer: A''' <math>\ddot{a}_{45}^{S}=1+v p_{45}^{S} \ddot{a}_{46}^{S U L T}</math> <math>p_{45}^{S}=e^{-\int_{0}^{1} \mu_{45+t}^{S} d t}=e^{-\int_{0}^{1}\left(\mu_{45+t}^{S U L T}+0.05\right) d t}=e^{-\int_{0}^{1}\left(\mu_{45+t}^{S U L T}\right) d t} e^{-\int_{0}^{1}(0.05) d t}=p_{45}^{S U L T} \cdot e^{-0.05}=\left(\frac{98,957.6}{99,033.9}\right) e^{-0.05}=0.9504966</math> <math>\ddot{a}_{45}^{S}=1+v p_{45}^{S} \ddot{a}_{46}^{S U L T}=1+(1.05)^{-1}(0.9504966)(17...")
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Exercise


Jan 18'24

Answer

Answer: A

[math]\ddot{a}_{45}^{S}=1+v p_{45}^{S} \ddot{a}_{46}^{S U L T}[/math]

[math]p_{45}^{S}=e^{-\int_{0}^{1} \mu_{45+t}^{S} d t}=e^{-\int_{0}^{1}\left(\mu_{45+t}^{S U L T}+0.05\right) d t}=e^{-\int_{0}^{1}\left(\mu_{45+t}^{S U L T}\right) d t} e^{-\int_{0}^{1}(0.05) d t}=p_{45}^{S U L T} \cdot e^{-0.05}=\left(\frac{98,957.6}{99,033.9}\right) e^{-0.05}=0.9504966[/math]

[math]\ddot{a}_{45}^{S}=1+v p_{45}^{S} \ddot{a}_{46}^{S U L T}=1+(1.05)^{-1}(0.9504966)(17.6706)=17.00[/math]

[math]100 \ddot{a}_{45}^{S}=1700[/math]

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