SOLUTION: If $39,500 is invested at 6.3% for 30 years, find the future value if the interest is compounded the following ways. (Round your answers to the nearest cent.)
(a) annually
$
Algebra.Com
Question 854689: If $39,500 is invested at 6.3% for 30 years, find the future value if the interest is compounded the following ways. (Round your answers to the nearest cent.)
(a) annually
$
Incorrect: Your answer is incorrect.
(b) semiannually
$
(c) quarterly
$
(d) monthly
$
(e) daily
$
(f) every minute (N = 525,600)
$
(g) continuously
$
(h) simple (not compounded)
$
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
If $39,500 is invested at 6.3% for 30 years, find the future value if the interest is compounded the following ways. (Round your answers to the nearest cent.)
the formula for future value is:
f = p * (1+i)^n where:
f = future value
p = present value
i = interest rate per time period
n = number of time periods.
time period can be yearly, semi-annually, quarterly, monthly, daily, every minute, or continuous.
if you are given the annual interest rate and the number of years, then you adjust your figures based on the time period used.
if the time period is annual, then the interest rate given per year and the number of years are used as is with no adjustment.
if the time period is quarterly, then the annual interest rate is divided by 4 and the number of years is multiplied by 4.
if the time period is monthly, then the annual interest rate is divided by 12 and the number of years is multiplied by 12.
essentially, you divide the annual interest rate by the number of compounding periods per year and you multiply the number of years by the number of compounding periods per year.
you will see this as we go through the individual cases.
(a) annually
$
annual interest rate is .063.
number of years is 30.
p = 39500.
no adjustment is necessary.
formula of f = p * (1+i)^n becomes f = 39500 * (1.063)^30 = 246942.04
(b) semiannually
$
annual interest rate is .063.
number of years is 30.
p = 39500.
annual interest rate is divided by 2 to get .0315.
number of time periods is equal to 2 * 30 = 60.
formula of f = p * (1+i)^n becomes f = 39500 * (1.0325)^60 = 253951.65
(c) quarterly
$
annual interest rate is .063.
number of years is 30.
p = 39500.
annual interest rate is divided by 4 to get .01575.
number of years is multiplied by 4 to get 120.
formula of f = p * (1+i)^n becomes f = 39500 * (1.01575)^120 = 257642.09
(d) monthly
$
annual interest rate is .063.
number of years is 30.
p = 39500.
annual interest rate is divided by 12 to get .00525.
number of years is multiplies by 12 to get 360.
formula of f = p * (1+i)^n becomes f = 39500 * (1.00525)^360 = 260175.58.
(e) daily
$
annual interest rate is .063.
number of years is 30.
p = 39500.
annual interest rate is divided by 365 to get .0001726027397.
number of years is multiplied by 365 to get 10950.
formula of f = p * (1+i)^n becomes f = 39500 * (1.0001726027397)^10950 = 261422.4238
(f) every minute (N = 525,600)
$
annual interest rate is .063.
number of years is 30.
p = 39500.
annual interest rate is divided by 525600 to get .0000001198630137.
number of years is multiplied by 525600 to get 15768000.
formula of f = p * (1+i)^n becomes f = 39500 * (1.0000001198630137)^15768000 = 261464.98
(g) continuously
$
annual interest rate is .063.
number of years is 30.
p = 39500.
a different formula is used here.
the formula is f = p * e^(in) where:
i = annual interest rate
n = number of years.
the formula of f = p * e^(in) becomes f = 39500 * e^(.063*30) = 261465.06
(h) simple (not compounded)
$
annual interest rate is .063.
number of years is 30.
p = 39500.
with simple interest there is no compounding.
the formula is f = p * i * n + p.
i is the annual interest rate
n is the number of years
the formula of f = p * i * n + p becomes f = 39500 * .063 * 30 + 39500 = 114155
the continuous compounding rate is the best that you can do.
as you increase the number of compounding periods per year, you approach the continuous compounding rate.
this is why you see progressively higher values when the compounding periods per year are increased.
the simple interest rate gives you the lowest value because there is no compounding.
RELATED QUESTIONS
If $33,500 is invested at 6.2% for 30 years, find the future value if the interest is... (answered by lenny460)
If $33,500 is invested at 6.1% for 30 years, find the future value if the interest is... (answered by kennydismuke)
If $3677 is invested at 6.14% for 4 years, find the future value if the interest is... (answered by ikleyn,MathTherapy)
1. If you lend $3600 to a friend for 15 months at 10% annual simple interest, find the... (answered by stanbon)
Find the future value of $2000 if it is invested for 4 years at an annual interest rate... (answered by Theo)
if 5000 is invested for t years at 6% interest, compounded monthly determine what the... (answered by rfer)
If $12,000 is invested at 4.5% for 20 years, find the future value of the interest is... (answered by ikleyn)
A couple bought some stock for $40 per share that pays an annual dividend of $0.80 per... (answered by Boreal,MathTherapy)
find the future value if 1000 if invested for 20 years at 8% compounded continously. (answered by jorel1380)