Answer: y=-3/2x+-8
Step-by-step explanation:
Ernie spend 6hrs. × his 3 sisters
so, 6×3=18
18hours in total.
It'd help if you could sketch this situation. Note that the area of a rectangle is equal to the product of its width and length: A = L W.
Consider the perimeter of this rectangular area. It's P = 2 L + 2 W. Note that P = 40 meters in this problem.
Thus, if we choose to use W as our independent variable, then P = 40 meters = 2 L + 2 W. Let's express L in terms of W. Divide both sides of the following equation by 2: 40 = 2 L + 2 W. We get 20 = L + W. Thus, L = 20 - w.
Then the area of the rectangle is A = ( 20 - W)*W.
Multiply this out. Your result will be a quadratic equation. Graph this quadratic equation (in other words, graph the function that represents the area of the rectangle). For which W value is the area at its maximum?
Alternatively, find the vertex of this graph: remember that the x- (or W-) coordinate of the vertex is given by
W = -b/(2a), where a is the coefficient of W^2 an b is the coefficient of W in your quadratic equation.
Finally, substitute this value of W into your quadratic equation, to calculate the maximum area.
Answer:
a) 
b) 0.0620
Step-by-step explanation:
We are given the following in the question:
Population mean, 
= 6
Variance, 
= 12
a) Value of 
We know that
Dividing the two equations, we get,
b) probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.
We can write the probability density function as:
We have to evaluate:
![P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620](https://tex.z-dn.net/?f=P%28x%20%3E12%29%5C%5C%5C%5C%3D%20%5Cdfrac%7B1%7D%7B16%7D%5Cdisplaystyle%5Cint%5E%7B%5Cinfty%7D_%7B12%7Df%28x%29dx%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B16%7D%5Cbigg%5B-2x%5E2e%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D-2%5Cdisplaystyle%5Cint%20xe%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7Ddx%7D%5Cbigg%5D%5E%7B%5Cinfty%7D_%7B12%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B8%7D%5Cbigg%5Bx%5E2e%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D%2B4xe%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D%2B8e%5E%7B-%5Cfrac%7Bx%7D%7B2%7D%7D%5Cbigg%5D%5E%7B%5Cinfty%7D_%7B12%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B8%7D%5Cbigg%5B%28%5Cinfty%29%5E2e%5E%7B-%5Cfrac%7B%5Cinfty%7D%7B2%7D%7D%2B4%28%5Cinfty%29e%5E%7B-%5Cfrac%7B%5Cinfty%7D%7B2%7D%7D%2B8e%5E%7B-%5Cfrac%7B%5Cinfty%7D%7B2%7D%7D%20-%28%20%2812%29%5E2e%5E%7B-%5Cfrac%7B12%7D%7B2%7D%7D%2B4%2812%29e%5E%7B-%5Cfrac%7B12%7D%7B2%7D%7D%2B8e%5E%7B-%5Cfrac%7B12%7D%7B2%7D%7D%29%5Cbigg%5D%5C%5C%5C%5C%3D0.0620)
0.0620 is the required probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.
