Revision as of 15:29, 19 November 2023 by Admin (Created page with "'''Solution: C''' <math display = "block">\begin{aligned} & 100(1+0.05)^2 \exp \left(\int_2^5 \frac{1}{1+t}\right)=100(1.1025) \exp \left[\left.\ln (1+t)\right|_2 ^5\right] \\ & =110.25\left(\frac{6}{3}\right)=220.50 \\ & \frac{100}{(1-d)^5}=220.50 \\ & (1-d)^5=\frac{100}{220.50}=0.453515 \\ & 1-d=0.853726 \\ & d=0.1463\end{aligned}</math> {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 19'23
Answer
Solution: C
[[math]]\begin{aligned} & 100(1+0.05)^2 \exp \left(\int_2^5 \frac{1}{1+t}\right)=100(1.1025) \exp \left[\left.\ln (1+t)\right|_2 ^5\right] \\ & =110.25\left(\frac{6}{3}\right)=220.50 \\ & \frac{100}{(1-d)^5}=220.50 \\ & (1-d)^5=\frac{100}{220.50}=0.453515 \\ & 1-d=0.853726 \\ & d=0.1463\end{aligned}[[/math]]
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