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| rev | Admin | (Created page with "'''Answer: E''' <math>L_{0}=100,000 v^{T}-560 \bar{a}_{\bar{T} \mid}=\left(100,000+\frac{560}{\delta}\right) e^{-\delta T}-\frac{560}{\delta}</math> Since <math>L_{0}</math> is a decreasing function of <math>T</math>, the <math>75^{\text {th }}</math> percentile of <math>L_{0}</math> is <math>L_{0}(t)</math> where <math>t</math> is such that <math>\operatorname{Pr}\left[T_{35}>t\right]=0.75</math>. <math>\frac{l_{35+t}}{l_{35}}=0.75</math> <math>l_{35+t}=0.75 l_{35}=...") | Jan 19'24 at 22:11 | +872 |