Answer: ![f^{-1}(x)=\frac{x}{5}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%3D%5Cfrac%7Bx%7D%7B5%7D)
Step-by-step explanation:
By definition the domain of an inverse function
is the range of f(x) and the range of the inverse function is equal to the domain of the principal function f(x).
If you have a function
, then to find the inverse function, follow these steps:
1. Make ![y=f(x)](https://tex.z-dn.net/?f=y%3Df%28x%29)
![f(x)=y=5x](https://tex.z-dn.net/?f=f%28x%29%3Dy%3D5x)
![y=5x](https://tex.z-dn.net/?f=y%3D5x)
2. Solve for the variable "x":
![x=\frac{y}{5}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7By%7D%7B5%7D)
3. Exchange the variable "x" with the variable "y":
![y=\frac{x}{5}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bx%7D%7B5%7D)
4. Exchange "y" with
. Then the inverse function is:
![f^{-1}(x)=\frac{x}{5}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%3D%5Cfrac%7Bx%7D%7B5%7D)
The coordinate (5,-6) lies in the fourth coordinate, because the first is x positive y positive, the second is x negative and y positive, the third is x negative and y negative, and the fourth is x positive y negative. This is x positive y negative, so it is in the fourth quadrant.
Answer:
Where
and ![\sigma=13.1](https://tex.z-dn.net/?f=%5Csigma%3D13.1)
We are interested on this probability
![P(X>140)](https://tex.z-dn.net/?f=P%28X%3E140%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule:
![P(z>1.924)=1-P(z](https://tex.z-dn.net/?f=P%28z%3E1.924%29%3D1-P%28z%3C1.924%29%3D0.0272)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and ![\sigma=13.1](https://tex.z-dn.net/?f=%5Csigma%3D13.1)
We are interested on this probability
![P(X>140)](https://tex.z-dn.net/?f=P%28X%3E140%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule:
![P(z>1.924)=1-P(z](https://tex.z-dn.net/?f=P%28z%3E1.924%29%3D1-P%28z%3C1.924%29%3D0.0272)
First plug in (x+h) for x in the function.
f(x+h)= 2(x-h)^2-3(x-h) = 2(x^2-2xh+h^2)-3x-3h =
2x^2-4xh+2h^2-3x-3h - 2x^2 +3x =
(-4xh +2h^2-3h)/h
-4x +2h-3