The probability density function of random variable x is f (x) = e ^ (- x), x > 0,0, X is less than or equal to 0, find e (x), e (2x), e (e ^ (- 2x))

The probability density function of random variable x is f (x) = e ^ (- x), x > 0,0, X is less than or equal to 0, find e (x), e (2x), e (e ^ (- 2x))

E(1)
Then E (x) = ∫ (- ∞, + ∞) f (x) xdx = 1
E(2X)=2E(X)=1
E(e^(-2X))= ∫(-∞,+∞)f(x)e^(-2x)dx=1/3
If you have any suggestions, you are welcome to discuss and learn together,

Let the density function of random variable X be FX = 2x (0

Classmate, the "density function" in your question should be "distribution function". You should read well! There are also some formulas in the formula book. The first step: find the density function of variable x, the method is to find the indefinite integral of F (x), the result is PX (x) = x ^ 2; the second step: find the inverse function of y = LNX, and find the derivative of the inverse function, the result is

Let the probability density of the random variable X be: FX (x) = 1 / 2 × e to the power of -- X, and find e (x). In the process of calculation, the absolute value of the negative absolute value of that e to the power of X
The absolute value of X is from negative infinity to positive infinity.

Needless to say, X * (1 / 2) e ^ (- |x |) is an odd function whose integral on the symmetric interval is 0
If x ^ 2 * (1 / 2) e ^ (- | x |), then it is equal to twice the positive half axis integral

When the plane parallel to the bottom of the cone is used to cut the cone, the ratio of the area of the cross section to the area of the bottom is 1:3. The ratio of the two sections which divide the generatrix of the cone into two sections is ()
A. 1：3B. 1：（3-1）C. 1：9D. 3：2

The plane parallel to the bottom of the cone is used to cut the cone. The truncated small cone is similar to the large cone. The ratio of the cross-sectional area to the bottom area is 1:3, so the similarity ratio of the small cone and the large cone is 1:3, so the generatrix length ratio of the small cone and the large cone is 1:3, so the generatrix length ratio of the small cone and the frustum is 1: (3-1), so B is selected

Let the density function of random variable X be f (x), f (- x) = f (x), and f (x) be the distribution function of X, then for any real number a, there is ()
A. F（-a）=1-∫a0f（x）dxB. F（-a）=12-∫a0f（x）dxC. F（-a）=F（a）D. F（-a）=2F（a）-1

F (- x) = f (x), from the definition, ∫ 0 − ∞ f (x) DX = 12, and because ∫ a0f (x) DX = - ∫ a0f (x) DXF (- a) = ∫ a − ∞ f (x) DX = ∫ 0 − ∞ f (x) DX + ∫ a0f (x) DX = 12 - ∫ a0f (x) DX, select: B

If a cone is cut by two planes parallel to the bottom and the side area ratio of the three cones is 1:2:3, the volume ratio of the three cones is?

You really have to emphasize the importance of "cone" in this question. That is to say, these three cones are 1) a single small cone, 2) a small cone + a middle cone of a small frustum, 3) a large cone of a small cone + a small frustum + a large frustum (that is, the original cone body)

Let the probability density of random variable X be φ (x), and φ (- x) = φ (x), and f (x) be the distribution function of X, then for any real number a, there is () --- f (- a) = 1 / 2 - ∫ [a, 0] φ (x) DX. The upper and lower limits of the integral are in brackets. How can I get this result,

As shown in the picture

Know the volume and bottom area of a cone, how to find the height of a cone?

The volume of a cone is one third of the volume of a cylinder with the same height and the same bottom!

Given the distribution function f (x) = a + B arctanx of random variable x, find P (- 1)

The distribution function is the integral of probability density over X. so p (- 1)

The volume of a cone is V, and the side area is s. at the middle point of its high line, it is cut by a section parallel to the bottom surface to form a cone and a frustum
(the frustum can also be regarded as a right angle trapezoid, which is formed by taking the straight line of the right angle waist as the rotation circle)
What is the volume and side area respectively?

1. Because it is truncated at the middle point of the high line, each segment accounts for half of the whole height. Therefore, it is deduced that the radius of the small circle is half of the radius of the large circle, so the area of the small circle is one fourth of the area of the large circle, and the volume of the small cone is 1 / 4 * 1 / 2 = 1 / 8V. 2