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...")
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]