Take a number x on the interval [- 1,2], then the probability of X ∈ [0,1] is______ ．

Take a number x on the interval [- 1,2], then the probability of X ∈ [0,1] is______ ．

On the number axis, the length of the line segment representing the interval [0,1] is 1; on the number axis, the length of the line segment representing the interval [- 1,2] is 3, so take a number x on the interval [- 1,2], then the probability of X ∈ [0,1] is P = 13, so the answer is: 13

∫∫ xdydz, where ∑ is the outside of the whole boundary surface of the space region bounded by z = x ^ 2 + y ^ 2 and z = 1

Can use Gauss formula directly!
2geh

In the interval [- 2,2], if two numbers x and y are taken randomly, then x ^ + y is satisfied^

Explain with geometric probability
It is the area ratio of unit circle area to rectangle [- 2,2] × [- 2,2]
That is, π / (4 × 4) = π / 16

Calculate the surface I = ∬ 2xz2dydz + y (z2 + 1) dzdx + (9-z3) DXDY, where ∑ is the surface z = x2 + Y2 + 1 (1 ≤ Z ≤ 2), take the lower side

Take plane ∑ 1: z = 2, take the upper side. Then ∑ and ∑ 1 form a closed surface, take the outer side. Let the space area surrounded by ∑ and ∑ 1 be Ω, and by Gauss formula, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; I = ∯∑ + ∑ 1 - ∬∑ 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp

In the interval [- 1,1], take a random number x, the probability of cos π x2 between 0 and 12 is ()
A. 13B. 2πC. 12D. 23

In order to make cos π x2 between 0 and 12, we need to make − π 2 ≤ π x2 ≤ − π 3 or π 3 ≤ π x2 ≤ π 2 ≠ − 1 ≤ x ≤ − 23 or 23 ≤ x ≤ 1, and the interval length is 23. The probability of cos π x2 between 0 and 12 is 232 = 13

Calculate the surface fraction I = ∫ (x ^ 3Z + X + Z) dydz - (x ^ 2yz + x) dzdx - (x ^ 2Z ^ 2 + 2Z) dzdx, where ∑ is the upper side of the surface z = 1-x ^ 2-y ^ 2 (Z ≥ 0)

In the region [- 1,1], the probability that the value of COS (π X / 2) is between 0 and 1 / 2 is removed randomly?

π X / 2 belongs to [- π / 2, π / 2]
In this range, cos at 0,1 / 2 is [- π / 2, - π / 3] ∪ [π / 3, π / 2]
One third of the total
So the probability is 1 / 3

Find the integral ∫ (x ^ 2 + ZX) dydz + (y ^ 2 + XY) dzdx + (Z ^ 2 + YZ) DXDY, where the integral is along the outside of the surface, x ^ 2 + y ^ 2 = Z ^ 2
What should we do after finding = ∫ ∫ (Z + X + y) dxdydz?

Is this cone not covered? The upper plane s: z = h, upper side ∫ ∫ (∑ + s) (X & ∫ + ZX) dydz + (Y & ∫ + 178; + XY) dzdx + (Z & ∫ + 178; + YZ) DXDY = ∫ ∫ Ω [(2x + z) + (2Y + x) + (2Z + y)] DV = 3 ∫ ∫ ∫ Ω (x + y + Z) DV, Ω are cone X & ∫ 178

Uniform distribution: the random variable x obeys the uniform distribution on the interval [0,0.2], and the probability density of X is calculated. Thank you

These two are about the same thing

1) What is the shape of the section when a plane parallel to the bottom is used to cut a cone? 2) what is the shape of the section when a plane perpendicular to the bottom is used to cut a cylinder?
Supplement: 2) is there the section with the largest area?

1) It looks like a circle
2) If it is a circle, because the cone is sharp at the top and thick at the bottom, it is the smallest at the top and the largest at the bottom