When P dollars is invested during seductiveness rate i, compounded annually, for t years, the investment grows to A dollars, where:
A=P(1-i)^t
a) Find the seductiveness rate i if $6250 grows to $6760 in 2 years.
Thank we so much. I will select the most appropriate answer tomorrow.
{ 4 comments… read them below or add one }
When you solve it you get i=-0.04. In the equation however it’s -(-0.04) so that’ still positive. I would say the answer is a rate of 0.04 or 4%.
6760 = 6250(1 + i)^2
6760 = 6250(i^2 + 2i + 1)
6760 = 6250i^2 + 12500i + 6250
6250i^2 + 12500i – 510
using the pythag. formula…..
it ends up being 4%
The formula is supposed to be:
A = P(1 + i)^t NOT A=P(1-i)^t
Given:
A = $6,760
P = $6,250
t = 2 yrs
Solution:
6760 = 6250(1 + i)^2
6250(1 + i)^2 = 6760
(1 + i)^2 = 6760/6250
1 + i = sqrt(6760/6250)
i = sqrt(6760/6250) – 1
i = 0.04
Therefore, the interest rate is 4%.
**Note: (1-i) should be (1+i).**
6760 = 6250(1+i)^2
Divide both sides by ‘6250′:
1.0816 = (1+i)^2
Take the natural logarithm of both sides:
ln1.0816 = 2ln(1+i)
Divide both sides by ‘2′:
(ln1.0816)/2 = ln(1+i)
Use both sides as a power to the exponential value (e):
e^((ln1.0816)/2) = e^(ln1+i)
which gives us:
1.04 = 1+i
Subtract ‘!’ from both sides:
0.04 = i
Therefore, the interest rate is 4%.
I hope this helps!