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| rev | Admin | (Created page with "'''Answer: A''' The present value random variable <math>\mathrm{PV}=1,000,000 e^{-0.05 T}, 2 \leq T \leq 10</math> is a decreasing function of <math>T</math> so that its <math>90^{\text {th }}</math> percentile is <math display="block"> \begin{aligned} & 1,000,000 e^{-0.05 p} \text { where } p \text { is the solution to } \int_{2}^{p} 0.4 t^{-2} d t=0.10 . \\ & \int_{2}^{p} 0.4 t^{-2} d t=-\left.0.4\left(\frac{t^{-1}}{-1}\right)\right|_{2} ^{p}=0.4\left(\frac{1}{2}-\f...") | Jan 18'24 at 12:32 | +595 |