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)
Answer: Obtuse
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They removed 4.5 tons in eight
The answer is c
This is because only the y-int changes
Answers:
right 1, down 3 (these two answers only)
Step-by-step explanation:
There are two transformations going on. The first of which is the horizontal or x translation (aka shifting) where the 'x' turns into 'x-1'. This moves the xy axis 1 spot to the left, giving the illusion that the graph moves 1 unit to the right.
The "-3" at the end shifts the graph down 3 units. If g(x) = f(x) - 3, then g(x) is the output of f(x) but 3 less than that, which is why it shifts down 3 units.
