Revision as of 13:35, 26 November 2023 by Admin (Created page with "'''Solution: B''' <math display="block"> 1200=1000 \exp \left(\int_0^4 \delta_r d r\right)=1000 \exp \left(\int_0^2 1.25 \delta d t+\int_2^4 \delta d t\right)=1000 \exp (2.5 \delta+2 \delta)=1000 \exp (4.5 \delta) </math> For the first year <math>1+i=e^{\int_0^1 1.25 \delta}=(1200 / 1000)^{1.25 / 4.5}</math> so <math>i=1.2^{1.25 / 4.5}-1=0.05195</math>. '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: B
[[math]]
1200=1000 \exp \left(\int_0^4 \delta_r d r\right)=1000 \exp \left(\int_0^2 1.25 \delta d t+\int_2^4 \delta d t\right)=1000 \exp (2.5 \delta+2 \delta)=1000 \exp (4.5 \delta)
[[/math]]
For the first year [math]1+i=e^{\int_0^1 1.25 \delta}=(1200 / 1000)^{1.25 / 4.5}[/math] so [math]i=1.2^{1.25 / 4.5}-1=0.05195[/math].
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.