Let the random variable x obey the uniform distribution in the interval (0, π), and calculate the probability density of the random variable y = - 2 ㏑ X

Let the random variable x obey the uniform distribution in the interval (0, π), and calculate the probability density of the random variable y = - 2 ㏑ X

FX (x) = 1 / π, X belongs to (0, π)
FX (x) = x / π, X belongs to (0, π)
FX(x)=-1.x>=π
FX(x)=0,x

Use your brain to think about it and do it by hand. Cut a cylinder with a plane. What shape is the cross section? A cone?

Circle, ellipse, rectangle (cylinder), triangle, ellipse, circle (cone)

Let X and y be independent random variables, and X obey uniform distribution in interval [0,1], and Y obey exponential distribution with parameter 1
(1) Finding the probability density of random variable z = x + y
(2) The density function of Z = - 2lnx

(1) It is known that f (x) = 1, (0

If you use a plane to cut a prism, a cone or a pyramid, what is the shape of the plane you can cut?

Triangle
A truncated prism cuts only one corner
The cone cuts from the apex
The pyramid also cuts from the vertex

Let X and y be independent of each other and obey the uniform distribution on [- 1,1], then e|x-y|=______ ．

The probability density function of uniform distribution x, y ~ U (- 1,1) is e (x) = ∫ baxb − ADX = 12 (a + b) = 0e (y) = 0; e | X-Y | = 0

What is the shape of a cylinder cut along its bottom diameter, and what is the shape of a cone cut along its top to bottom diameter

A cylinder is cut along the diameter of its base and its section is a rectangle or square
A cone is cut along the diameter from the top to the bottom with a triangular section

Let x obey the uniform distribution in the interval [0,1], y obey the exponential distribution with parameter 1, and X and y are independent of each other

fx(x)=1,fy(y)=e^-y
fx,y(x,y)=fx(x)fy(y)=e^-y
P(x>y)=P(x>y|Y=y)=1-P(x

If you cut a cube with a plane, you can't get a right triangle. Why

Each side of the cut triangle forms a right triangle with the two sides of the cube. Obviously, the two sides of the square are right angles. Let the three right angles be a, B and C respectively, then the square of the side length of the cross-section triangle can be expressed as a ^ 2 + B ^ 2, B ^ 2 + C ^ 2, C ^ 2 + A ^ 2
In the above three groups of numbers, any two can not be equal to the third group, so it can not be a right triangle

Simple probability theory and mathematical statistics, to determine which can be used as a random variable distribution function, which can not, why
 

(3) (4) both are OK, because ① 0 ≤ f (x) ≤ 1; and; f (- ∞) = 0; f (+ ∞) = 1; ② f (x) is a nondecreasing function; ③ f (x) is right continuous

Try it: can you get an equilateral triangle by cutting a cube with a plane? Can you cut a right triangle or an obtuse triangle?
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No
The reasons are as follows.
First of all, cut a cube with a plane. To get a triangle, it must intersect with three adjacent planes (as shown in the figure). The following only needs to explain that the section △ PMN is not a right triangle or an obtuse triangle
There are three right triangles, △ BMN, △ PMB and △ PBN in the graph. If Mn is the longest side, then we only need to explain that PM2 + pN2 and Mn2 are not equal
In RT △ BMN, according to Pythagorean theorem, BM2 + BN2 = Mn2;
Similarly, BM2 + bp2 = MP2, BN2 + bp2 = NP2. That is, PM > BM, PN > BN
Therefore, PM2 + pN2 > Mn2. From the cosine theorem, △ PMN is an acute triangle