U2 = {(x1, x2, x3, x4) ∈ R 4 | x1 + x2 + x3 − x4 = x1 − 2x2 + x4 = 0} My attempt: for U1; I created a vector in which one variable, different in each vector, is zero and another is 1 and got three vectors: (3,0,-1,1), (0,3,-2,1), (2,1,0,1)

Hi I have this question in my book (preparing for the exam) and I can't seem to find the answer.. X1 and X2 are independent and identically distributed. They are continuous and uniform over [-1,1]...

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2018年4月24日 · Let random variable X1, with variance one, has the following property: $\frac {X1+X2} {\sqrt2}$ has the same distribution as X1, where X2 is an independent copy of X1. Show that X1 ∼ N(0, 1). So i have to show x1 is normal distribution; its very easy to show the other way of course, when x1 is normal distirbution, then obviously $\frac {X1+X2 ...

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2017年10月15日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

where x= \begin{bmatrix}x1\\x2\\x3\\x4\end{bmatrix} Find a matrix A such that T(x)=A(dot)x. Then verify the equality T(x)=A(dot)x. As an answer, would A = \begin{bmatrix}2&-3&0&0\\5&0&6&0\\7&0&0&-8\end{bmatrix} If my A is correct, when i try to verify the equality T(x)=A(dot)x would i just get \begin{bmatrix}2x1-3x2\\5x1+6x3\\7x1-8x4\end ...

2017年9月18日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Say we have these two points on a xy plane. (2, 3) and (4, 5). The slope would be 1. To get the equation of the line we can use one of these two formulas. Substitution y and x values in the formul...

2017年5月16日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.