Recovery Disk Windows Vista Home Premium X14 39682
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Nevertheless, if you have the product key, you can get the product from the following place:
Then you can check the size with “Microsoft Windows operating system version history”
If the disc is OEM, it will be installed by Microsoft as OEM, which means that the product is not legally available, and you have a legal problem if you try to use it on your computer. Because, the disk is like a black box, it is not ready to be used by any one of your end users.
And that’s how you can get the microsoft manufacturer original disk.
If you want the recovery disk, contact the product manufacturer and ask them if there is an OEM copy for your hardware. If not, you will only get the retail copy.
But in the case of windows 7, they changed the status of OEM for all the new versions. And they have back to OEM for all the OEM editions, but not for retail editions. So you can only get OEM for the retail version.
Q:
Finding the expected value of $\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+\frac{x}{1!}$
You have a fair normal distribution with mean $μ=2$ and standard deviation $σ$.
Now you are asked to find the expected value of the following sum:
$\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+\frac{x}{1!}$
I saw this in a contest a few months ago, but I don’t really know how to do this. All I can think of is:
$\frac{1}{σ\sqrt{2\pi}}\int_{ -\infty}^{+\infty} e^{\frac{(x-μ)^2}{2σ^2}}\cdot e^{\frac{ -(x-μ)^2}{2}}\,\mathrm dx$
That is, I’m using the fact that $E(X) = \int x\cdot f(x)\,\mathrm dx$
Also, I did solve