what formula would you use for this problem?
2007-04-03 14:24:11 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Investing $1000 for 8 years at 8% assuming annual compounding. (1.08)^8 x 1000 = 1850.83
At the end of year:
1 - $1080.00
2 - $1166.40
3 - $1259.71
4 - $1360.49
5 - $1469.33
6 - $1586.87
7 - $1713.82
8 - $1850.93
9 - $1999.00
I added a ninth year because of something known as the "rule of 72". Take your interest rate and divide it into 72. That is the number of years it will take your money to double. 72 / 8 = 9. In nine years, your investment doubled to $2000.
2007-04-03 14:28:14 · answer #1 · answered by Thomas C 6 · 1⤊ 0⤋
You raise the rate per period added to one to the power of the number of period.
Rate is 8% (or 0.08) add 1 and you have 1.08.
Raise that to the 8th power (i.e. multiply 1.08 by itself 8 times) and you get 1.85093.
So, $1 will be worth $1.85 or so; $1000 will be worth $1850.93
2007-04-03 14:30:28 · answer #2 · answered by Vincent G 7 · 1⤊ 0⤋
8 % is 0.08
every year you multiply the amount by 1.08
In n years, you would have 1000* (1.08^n)
in 8 years 1000*(1.08^8) =1000*1.85093021 or 1850.93
2007-04-03 15:23:33 · answer #3 · answered by PC_Load_Letter 4 · 0⤊ 0⤋
$1,850.93
Just plug in the numbers in to the intrest calculator and it solves for you.
2007-04-03 14:34:37 · answer #4 · answered by Mr. Smith 5 · 0⤊ 0⤋