Revision as of 22:18, 19 November 2023 by Admin (Created page with " '''Solution: D.''' <math display = "block"> \begin{aligned} & P(0, m)=(1+i)^{-m} \\ & P(0, n)=(1+i)^{-n} \\ & X=\frac{(1+i)^{-m}}{(1+i)^{-n}}=(1+i)^{-m+n} \\ & X=\frac{P(0, m)}{P(0, n)}\end{aligned} </math> {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 19'23
Answer
Solution: D.
[[math]]
\begin{aligned} & P(0, m)=(1+i)^{-m} \\ & P(0, n)=(1+i)^{-n} \\ & X=\frac{(1+i)^{-m}}{(1+i)^{-n}}=(1+i)^{-m+n} \\ & X=\frac{P(0, m)}{P(0, n)}\end{aligned}
[[/math]]
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