Revision as of 00:27, 26 November 2023 by Admin (Created page with "'''Solution: E''' We are given <math>i^{(4)}=.06</math>, so <math>1+i=\left(1+\frac{i^{(4)}}{4}\right)^4=1.06136</math>. The first deposit occurs at time 0 , the second at time 0.5 . We need to calculate the amount accumulated at time 2.5. The first deposit accumulates to <math>100(1+i)^{2.5}=116.054</math>, the second deposit to <math>200(1+i)^2=225.2985</math>. The sum is 341.35 . '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: E
We are given [math]i^{(4)}=.06[/math], so [math]1+i=\left(1+\frac{i^{(4)}}{4}\right)^4=1.06136[/math]. The first deposit occurs at time 0 , the second at time 0.5 . We need to calculate the amount accumulated at time 2.5. The first deposit accumulates to [math]100(1+i)^{2.5}=116.054[/math], the second deposit to [math]200(1+i)^2=225.2985[/math]. The sum is 341.35 .
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.