Question

suppose you have $3000 to invest. which investment yields the greater return over 10 years: 6.5% compounded semiannually or 6% compounded monthly? how much more is yielded by the better investment?

Answer 1

Well 10*.065 because you move the decimal two places to the left equals .65 add that to your 3000. then do your 10*.06 to get .6 then add it to your 300 i think that is how you would do this type of problem. Does it have any answers to it that you have to choose from?

Answer 2

unless you would times your investment 3000 by .065 & then .06. 3000+195(10)=4950 and the second one 3000+180(10)=4800.

Answer 3

\[Use the equation A=P(1+r/n)^{t*n} Where P is the amount invested r=interest rate, n is the amount of \times the interest is done a year(quarterly, semianually), t is the time invested for\]

Answer 4

Use that equation P= amount invested, r=interest rate, n=amount of times the interest is compounded in(quarterly or semiannually), t is amount of time money is invested.

Answer 5

Compound Interest formula:\[\Large A = P(1 + \frac{ r }{ n })^{nt}\]
P = Principal
A = Amount at maturity (or the balance on the account)
r = annual rate of interest in decimal
n = compounding period
t = number of years

For both cases, P = 3000, t = 10

For 6.5% compounded semiannually, r = 0.065, n = 2 (semi-annual means twice a year)
Find A.

For 6% compounded monthly, r = 0.06, n = 12
Find A

Compare and see which one gives higher A.

Answer 6

@shannonb0284: To keep text from bunching together when using the Equations editor try:

\text { Whatever you want to stay without the words bunching up } :
\[\text { Whatever you want to stay without the words bunching up } \]