Suppose that on January 1, 2018, you buy a bond for $2,000 that will pay interest of 3.6% per year compounded continuously for 2
1 answer:
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Answer:
a.$7,096
Step-by-step explanation:
- total amount A = s(\frac{(1+\frac{r}{n}) ^{n*t}-1 }{\frac{r}{n} } )[/tex]
[this is the formula used when certain amount is added at regular intervals and if the interest is COMPOUNDED]
(here,
s=deposit added at the end of each quarter= $500
n= number of times interest is compounded per unit 't'=4
t=total time = 3 years
r=interest rate ( expressed in decimal or fraction not in %) = 12/100 = 0.12 )
- A= 500(\frac{(1+\frac{0.12}{4}) ^{4*3}-1 }{\frac{0.12}{4} } )[/tex]
- by solving further,
A
( [1.03]^12= 1.42576 ( approximately) { use calculator for exact values})