Respuesta :
Answer:
a,b,c,d,e,f) The effective annual yield is 1.1P-P = 0.1P = 10%P.
Step-by-step explanation:
This is a compound interest problem
Compound interest formula:
The compound interest formula is given by:
[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]
A: Amount of money(Balance)
P: Principal(Initial deposit)
r: interest rate(as a decimal value)
n: number of times that interest is compounded per unit t
t: time the money is invested or borrowed for.
a)
r = 0.06
n: 2(semianually means that the interest is compounded twice a year).
t = 1.
[tex]A = P(1 + \frac{0.06}{2})^{2*1}[/tex]
[tex]A = 1.1P[/tex]
The acount started the year with P, and it ended with 1.1P, so the effective annual yield is 1.1P-P = 0.1P = 10%P.
b)
Now we have n = 3, since if the interest is compounded quarterly, is is compounded three times a year(a year has 3 quarters). So:
[tex]A = P(1 + \frac{0.06}{3})^{3*1}[/tex]
[tex]A = 1.1P[/tex]
The effective annual yield is 1.1P-P = 0.1P = 10%P.
c) Now we have n = 12, since the interest is compounded monthly, and there are 12 months a year. So:
[tex]A = P(1 + \frac{0.06}{12})^{12*1}[/tex]
[tex]A = 1.1P[/tex]
The effective annual yield is 1.1P-P = 0.1P = 10%P.
d) Since the interest is compounded daily, and we assume 360 days in a year, n = 360. So:
[tex]A = P(1 + \frac{0.06}{360})^{360*1}[/tex]
[tex]A = 1.1P[/tex]
The effective annual yield is 1.1P-P = 0.1P = 10%P.
e) The interest is compounded 1000 times a year, so n = 1000
[tex]A = P(1 + \frac{0.06}{1000})^{1000*1}[/tex]
[tex]A = 1.1P[/tex]
The effective annual yield is 1.1P-P = 0.1P = 10%P.
f) The interest is compounded 100000 times a year, so n = 100000
[tex]A = P(1 + \frac{0.06}{100000})^{100000*1}[/tex]
[tex]A = 1.1P[/tex]
The effective annual yield is 1.1P-P = 0.1P = 10%P.
