Revision as of 13:38, 26 November 2023 by Admin (Created page with "What is the value of <math display = "block"> \left(1+\frac{i^{(m)}}{m}\right)(1-d)^{1 / m} ? </math> <ul class="mw-excansopts"><li><math display = "block">(1+i)^{2 / m}</math></li><li><math display = "block">\frac{i^{(m)} d^{(m)}}{m}</math></li><li><math display = "block">1-\frac{i^{(m)} d}{m}</math></li><li><math display = "block">\left(1-\frac{d^{(m)}}{m}\right)^2</math></li><li><math display = "block">1 </math></li></ul> '''References''' {{cite web |url=https://w...")
ABy Admin
Nov 26'23
Exercise
What is the value of
[[math]]
\left(1+\frac{i^{(m)}}{m}\right)(1-d)^{1 / m} ?
[[/math]]
- [[math]](1+i)^{2 / m}[[/math]]
- [[math]]\frac{i^{(m)} d^{(m)}}{m}[[/math]]
- [[math]]1-\frac{i^{(m)} d}{m}[[/math]]
- [[math]]\left(1-\frac{d^{(m)}}{m}\right)^2[[/math]]
- [[math]]1 [[/math]]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.
ABy Admin
Nov 26'23
Solution: E
Let A be the answer. Then
[[math]]
A^m=\left(1+\frac{i^{(m)}}{m}\right)^m(1-d)=(1+i)(1-d)=(1+i) v=1 .
[[/math]]
So [math]\mathrm{A}=1[/math].
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.