Revision as of 23:59, 25 November 2023 by Admin (Created page with "'''Solution: E''' <math display="block"> \begin{aligned} \text { Accum.Value } & =1000 e^{\int_0^5 \delta_t d t}+1000 e^{\int_2^5 \delta_t d t}+1000 e^{\int_4^5 \delta_t d t} \\ & =1000 e^{(\ln (1.05))(25 / 2-0)}+1000 e^{(\ln (1.05))(25 / 2-4 / 2)}+1000 e^{(\ln (1.05))(25 / 2-16 / 2)} \\ & =1000\left(1.05^{12.5}+1.05^{10.5}+1.05^{4.5}\right)=1000(1.840205+1.669120+1.245523)=4754.85 \end{aligned} </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/...")
Exercise
ABy Admin
Nov 25'23
Answer
Solution: E
[[math]]
\begin{aligned}
\text { Accum.Value } & =1000 e^{\int_0^5 \delta_t d t}+1000 e^{\int_2^5 \delta_t d t}+1000 e^{\int_4^5 \delta_t d t} \\
& =1000 e^{(\ln (1.05))(25 / 2-0)}+1000 e^{(\ln (1.05))(25 / 2-4 / 2)}+1000 e^{(\ln (1.05))(25 / 2-16 / 2)} \\
& =1000\left(1.05^{12.5}+1.05^{10.5}+1.05^{4.5}\right)=1000(1.840205+1.669120+1.245523)=4754.85
\end{aligned}
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.