What is the probability that the sum of any two real numbers in the closed interval [- 1,1] is not greater than 1

What is the probability that the sum of any two real numbers in the closed interval [- 1,1] is not greater than 1

7/8
As long as x + y = 1 intersects with X axis and Y axis, the area above the straight line is 1 / 2, and the area of the whole interval is 2 * 2 = 4.1 / 2 / 4 = 1 / 8, so the probability of no more than 1 is 7 / 8

Geometric probability type: if any real number x is taken in the interval [0,10], then the probability of real number x not greater than 2 is?

The large interval is divided into five equal small intervals, i.e. [0,2], (2,4], (4,6], (6,8], (8,10], the probability of not more than 2 is of course 1 / 5

If the product of (x ^ 2 + PX + Q) (x ^ 2-3x + 2) does not contain x ^ 2 and x ^ 3 terms, find PQ

A:
After expansion, the items with x ^ 3 and x ^ 2 are as follows:
-3x*x^2+x^2*px=(p-3)x^3；
qx^2-3x*px+2x^2=(q+2-3p)x^2.
Therefore, the equations are solved
p-3=0；
q+2-3p=0
The solution is p = 3, q = 7

If a = {x | x ^ + PX + q = 0, X ∈ r}, B = {x | x ^ - 3x + 2 = 0, X ∈ r}, a ∪ B = B, find the conditions that P and Q satisfy

B (x-1)(x-2)=0
x=1,x=2
So B = {1,2}
A∪B=B
So a is a subset of B
There are three situations
(1) A is an empty set
Then the equation has no solution
So the discriminant p ^ 2-4q

Set a = {x | x ^ 2 + PX + q = 0}
Given a = {X / x ^ 2 + PX + q = 0}, B = {X / QX ^ 2 + PX + 1 = 0}, and satisfying a ∩ B = empty set, - 2 belongs to a, find the value of P and Q

-2 belongs to a
So: q-2p + 4 = 0
For a set: delta = P ^ 2-4q = P ^ 2-8p + 16
It seems that we can only work out the range, not the specific value

Given that P belongs to R, set a = {x | x ^ 2-px-5 / 2 = 0}, set B = {x | x ^ 2-9 / 2 ^ 2-P = 0}, if 1 / 2 belongs to a, find all elements in B

∵1/2∈{x|x^2-px-5/2=0},
∴（1/2)^2-p/2-5/2=0
The solution is as follows
p=-9/2
Then the set {x | x ^ 2-19 / 2 ^ 2 + 9 / 2 = 0} is:
{x|x^2-1/4=0}
The results are as follows
x=±1/2
That is, the elements in the set are: {x | - 1 / 2,1 / 2}

The known set a = {x │ x ^ 2-2x-3 ≤ 0}, B = {x │ x ^ 2 + PX + Q}

A medium
xx - 2x -3
-1

If the solution set of inequality 2x + PX + Q ＞ 0 is {x | x ＜ - 1 or X ＞ 3},

The quadratic equation of one variable is 2x ^ 2 + PX + q = 0, two of which must be - 1 and 3, then - 1 + 3 = - P / 2, P = - 4 - 1 * 3 = q / 2, q = - 6, PQ = 24

Given the set a = {x | x + PX + q = 0} = {2}, find the value of P + Q + PQ
Please describe in detail how to find the value of P and Q

The set has only one element 2, which means that the equation x + PX + q = 0 has only one root. If it is 2, then 4 + 2p + q = 0 and p-4q = 0, so p = - 4, q = 4, P + Q + PQ = 16

Given {x | X's square + PX + q = 0} = {2}, find the value of P's square + Q's square + PQ
The relationship between root and coefficient shows that 2 + 2 = - P, 2 × 2 = Q

The relationship between root and coefficient
Sum of two = - P, product of two = q
And {the square of X | x + PX + q = 0} = {2},
It shows that the equation has two equal real roots 2
SO 2 + 2 = - P, 2 × 2 = Q