Revision as of 23:29, 26 November 2023 by Admin (Created page with "'''Solution: A''' <math display = "block"> \begin{aligned} & 100 a_{\overline{60} \mid .01}=6000 v^K \text { so } \\ & 100 \frac{1-1.01^{-60}}{.01}=6000(1.01)^{-K} \text { so } \\ & K=-\left(\ln (10 / 6)+\ln \left(1-1.01^{-60}\right)\right) / \ln (1.01)=29.01227 .\end{aligned} </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor.ca | title = University of Windsor Old Tests 62-...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: A
[[math]]
\begin{aligned} & 100 a_{\overline{60} \mid .01}=6000 v^K \text { so } \\ & 100 \frac{1-1.01^{-60}}{.01}=6000(1.01)^{-K} \text { so } \\ & K=-\left(\ln (10 / 6)+\ln \left(1-1.01^{-60}\right)\right) / \ln (1.01)=29.01227 .\end{aligned}
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.