SOLUTION: Please Help Me Solve:
An investment has been growing at a rate of 8% per year and now has a value of $ 8500. What was the value of the investment, to the nearest dollar, 4 years a
Algebra.Com
Question 31711: Please Help Me Solve:
An investment has been growing at a rate of 8% per year and now has a value of $ 8500. What was the value of the investment, to the nearest dollar, 4 years ago.
Using A=P(1+t)r
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
AN ALTERNATE FORMULA IS INTEREST OF AN INVESTMENT FOR 4 YEARS @ 8% IS
I=4X.08 OR I=.32 OR ORIGINAL INVESTMENT TIMES 1.32I=$8,500 OR I=$8,500/1.32 OR I=$6,439
PROOF $6,439+$6,439(4).08=$8,500 OR $6,439+$6,439(.32)=$8,500 OR $6,439+$2,060=$8,499
RELATED QUESTIONS
an avestment is losing value at a rate of 8% per year and now has a value of $8200. find... (answered by stanbon)
An investment has grown to $89,000 over a 6-year period. The interest was computed daily (answered by checkley77)
The city of Anville is currently home to 28000 people, and the population has been... (answered by ewatrrr,greenestamps)
If $8500 is invested at a rate of 6% per year compounded daily, what is the balance after (answered by nerdybill)
Precision Machining corporation has been growing steadily over the past decade. Demand... (answered by ikleyn)
Skip to content
Using Gmail with screen readers
devnull@chudov.com
6 of 11... (answered by Alan3354)
Jenny receives $1270 per year from three different investments totaling $20,000. One of... (answered by solver91311)
According to the 2011 census, the population of Ottawa was 883390. The population
has... (answered by ikleyn)