I met to type in school subject.
A. A=P(1+r/n)^n*t
A=$3000(1*0.02/8)^8*8
A=$3000(1+0.0025)^64
A=$3000(1.0025)^64
A=3000(1.173276583)
A=3519.829749=3520.82
B. A=P(1+r/n)^n*t
2000=$3000(1+0.02/8)^8*t
-2000-2000
$1000(1+0.00025/8)^8t
$1000(1.0025)8t
1000(1.020175878)
=1020.175878=102.185
This problem practice test:
Abe wants to invest $3000 in a bond paying 2% interest.
A. What will be the amount of the bond after it matures in 8 years.
B. How long will it take to double?
2 answers
You have the correct formula
A=P(1+r/n)^(n*t)
but "n" is the number of times per year that the interest is compounded. Since it does not specify a shorter compounding period, you should assume annual compounding, so n=1. So, you should proceed with
A. 3000(1+0.02)^8 = 3514.98
B. 3000(1+.02)^t = 2*3000
Not sure where you got that 2000, since it bears no relation to doubling. In fact, the doubling time is unrelated to the amount invested, since the 3000 can be canceled, leaving
(1+.02)^t = 2
t log1.02 = log2
t = log2/log1.02 = 35
A=P(1+r/n)^(n*t)
but "n" is the number of times per year that the interest is compounded. Since it does not specify a shorter compounding period, you should assume annual compounding, so n=1. So, you should proceed with
A. 3000(1+0.02)^8 = 3514.98
B. 3000(1+.02)^t = 2*3000
Not sure where you got that 2000, since it bears no relation to doubling. In fact, the doubling time is unrelated to the amount invested, since the 3000 can be canceled, leaving
(1+.02)^t = 2
t log1.02 = log2
t = log2/log1.02 = 35
0 answers