Question 47662This question is from textbook College Algebra
: Which of the given interest rates and compounding periods would provide the better investment?
(a) 9 1/4% per year, compounded semiannually
(b) 9% per year, compounded continously
Thank you
This question is from textbook College Algebra
Found 2 solutions by stanbon, venugopalramana:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Consider the result after one year (t=1)
(a) 9 1/4% per year, compounded semiannually
A=P(1+0.0925/(2t)^(2t)
A=P(1+0.04625)^2
A=P(1.09463906...
(b) 9% per year, compounded continously
A=Pe^(0.09t)
A=Pe^0.09
A=P(1.09417428...)
So,for one year the 9 % compounded continuously is growing faster.
You would have to look at other values of t (maybe by graphing)
to see which is better in the long run. I suspect the
continuous compounding would be the winner.
To do this, let P be $1.00 and let t=x.
Cheers,
Stan H.
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! Which of the given interest rates and compounding periods would provide the better investment?
FORMULA IS
A=P{1+R/100N}^NT
(a) 9 1/4% per year, compounded semiannually
R=9.25....T=??..TAKEN AS 1 YEAR...N=2
A=P(1+9.25/200)^2=1.09464P
(b) 9% per year, compounded continously
R=9........T=1......N=TENDS TO INFINITY
A=P(1+9/100N)^N...LIMIT N TENDING TO INFINITY..
=P(1+1/X)^X}^9/100.....WHERE X=100N/9
=P(E)^(9/100)=1.09417
HENCE 9 1/4% AT 6 MONTHS COMPOUNDING GIVES HIGHER RETURN
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