I = p * r * n
i is the interest
p is the principal
r is the interest rate per time period
n is the number of time periods.
in your problem:
i = 900
p = 2000
r = what you want to find
n = 3 years
formula becomes 900 = 2000 * r * 3
solve for r to get r = 900 / 2000 / 3 = .15
that's .15 interest rate per year = 15% per year.
at a nominal interest rate of .15 per year, the interest rate per month would be .15/12 = .0125 per month.
the remaining balance at the end of 6 month is equal to 1907.140183
Answer:
where is the figure?
Step-by-step explanation:
I can't see a figure
Answer:
<h2>7</h2>
Step-by-step explanation:
![\left[\left(11\:-\:4\right)^3\right]^2\:\div \left(4\:+\:3\right)^5\\\\\frac{\left(\left(11-4\right)^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Subtract\:the\:numbers:}\:11-4=7\\\\=\frac{\left(7^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Add\:the\:numbers:}\:4+3=7\\\\=\frac{\left(7^3\right)^2}{7^5}\\\\\left(7^3\right)^2=7^6\\\\=\frac{7^6}{7^5}\\\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\\frac{7^6}{7^5}=7^{6-5}\\\\\mathrm{Subtract\:the\:numbers:}\:6-5=1\\\\=7](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft%2811%5C%3A-%5C%3A4%5Cright%29%5E3%5Cright%5D%5E2%5C%3A%5Cdiv%20%5Cleft%284%5C%3A%2B%5C%3A3%5Cright%29%5E5%5C%5C%5C%5C%5Cfrac%7B%5Cleft%28%5Cleft%2811-4%5Cright%29%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A11-4%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BAdd%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A4%2B3%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cleft%287%5E3%5Cright%29%5E2%3D7%5E6%5C%5C%5C%5C%3D%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Aexponent%5C%3Arule%7D%3A%5Cquad%20%5Cfrac%7Bx%5Ea%7D%7Bx%5Eb%7D%3Dx%5E%7Ba-b%7D%5C%5C%5C%5C%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%3D7%5E%7B6-5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A6-5%3D1%5C%5C%5C%5C%3D7)
I think it's the other way around. A quantity that doubles for each additional unit of time is an exponential growth function. An example would be
kt
N = N 2
o
where k is the "growth constant."
