Revision as of 17:43, 26 November 2023 by Admin (Created page with "'''Solution: E''' <math>100=a(0)=M .110=a(1)=K+L+M</math> so <math>K+L=10.136=a(2)=4 K+2 L+M</math> so <math>18=2 K+L</math> so <math>K=8</math> and <math>L=2</math>. Thus <math>a(t)=8 t^2+2 t+100</math>. So <math>\delta(t)=\frac{a^{\prime}(t)}{a(t)}=\frac{16 t+2}{8 t^2+2 t+100}</math> so <math>\delta(.5)=\frac{10}{103}=.097</math>. '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor.c...")
Exercise
ABy Admin
Nov 26'23
Answer
Solution: E
[math]100=a(0)=M .110=a(1)=K+L+M[/math] so [math]K+L=10.136=a(2)=4 K+2 L+M[/math] so [math]18=2 K+L[/math] so [math]K=8[/math] and [math]L=2[/math]. Thus [math]a(t)=8 t^2+2 t+100[/math]. So [math]\delta(t)=\frac{a^{\prime}(t)}{a(t)}=\frac{16 t+2}{8 t^2+2 t+100}[/math] so [math]\delta(.5)=\frac{10}{103}=.097[/math].
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.