Exercise
ABy Admin
Nov 26'23
Answer
Solution: C
[[math]]A_X(t)=1 e^{\int_0^t \delta_r d r}=e^{.005 t^2+.1 t}, \, A_Y(t)=(1+i)^t[[/math]]
Thus [math]e^{.005(20)^2+.1(20)}=e^4=[/math] set [math]=(1+i)^{20}[/math]. We want to find
[[math]]
(1+i)^{1.5}=\left((1+i)^{20}\right)^{1.5 / 20}=\left(e^4\right)^{1.5 / 20}=e^{.3}=1.34986
[[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.