You can compute both the mean and second moment directly using the density function; in this case, it's
Then the mean (first moment) is
![E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x\,\mathrm dx=710](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7B80%7D%5Cint_%7B670%7D%5E%7B750%7Dx%5C%2C%5Cmathrm%20dx%3D710)
and the second moment is
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\frac1{80}\int_{670}^{750}x^2\,\mathrm dx=\frac{1,513,900}3](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7B80%7D%5Cint_%7B670%7D%5E%7B750%7Dx%5E2%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7B1%2C513%2C900%7D3)
The second moment is useful in finding the variance, which is given by
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2=\dfrac{1,513,900}3-710^2=\dfrac{1600}3](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2%3D%5Cdfrac%7B1%2C513%2C900%7D3-710%5E2%3D%5Cdfrac%7B1600%7D3)
You get the standard deviation by taking the square root of the variance, and so
![\sqrt{V[X]}=\sqrt{\dfrac{1600}3}\approx23.09](https://tex.z-dn.net/?f=%5Csqrt%7BV%5BX%5D%7D%3D%5Csqrt%7B%5Cdfrac%7B1600%7D3%7D%5Capprox23.09)
<span>4x-6y=24
2x-3y=12
2x-12=3y
y=2/3*x-4</span>
Answer:
x = 1/4
Step-by-step explanation:
16 = 2 can be written as 
= 2
(
)^x = 2^1
if bases are the same we can create this equation: 4x = 1
x = 1/4
Answer:
see attachment (this needs to be at least 20 characters long)
Answer:
im not sure if this is right or not but you can try it if you want
Step-by-step explanation:
A. 75.8 square units B. 76.4 square units C. 76.8 square units D. 79.4 square units
