Let all possible values of continuous random variable X be in the interval [a, b], and its density function be f (x)

Let all possible values of continuous random variable X be in the interval [a, b], and its density function be f (x)

Hungry A simplified version of last semester's assignment on probability theory The problem I did didn't tell x is continuous, and it can also prove these two conclusions. I'll write down the standard method that the teacher said. ① a ≤ x ≤ B, the expectation E has order preservation, which is a theorem. So e (a) ≤ e (x) ≤ e (B), and then the expectation of constant is of course equal to this

If the density function of random variable x is symmetric with respect to x = u, it is proved that its distribution function satisfies the following requirements: F (U + x) + F (u-x) = 1 (x is between positive and negative infinity)
Let the density function be f (x), f (U-T) = f (U + T), and t be all real numbers
F (U + x) = ∫ (upper limit U + X, lower limit negative infinity) f (s) ds = ∫ (upper limit x, lower limit negative infinity) f (U + T) DT
I'm not very clear about the upper and lower limit integral transformation of the above step

If t = u + X, then x = U-T DX = - DT

Let the density function f (x) of the random variable X be a continuous function and its distribution function f (x), then 2F (x) f (x) is a probability density function

Let g (x) be a new distribution function
G (x) = ∫ (- infinity, x) 2F (T) f (T) DT = ∫ 2F (T) DF (T) = [f (x)] ^ 2 continuous form is not a problem,
Limg (x) = Lim [f (x)] ^ 2 = 1 (x - > positive infinity)
Limg (x) = Lim [f (x)] ^ 2 = 0 (x - > negative infinity)
So g (x) satisfies the distribution function, and its corresponding density function is 2F (x) f (x)

What is 326 minus 58 minus 2

326-58-2
=326-（58+2）
=326-60
=266

Given that the line y = KX + B and the line y = 2x-3 intersect at the same point on the Y-axis and pass through the point (m, 6) on the line y = - 3x, the analytical formula is obtained

From the solution of the problem y = KX + B and y = 2x-3 intersect at (0, - 3), y = - 3x intersect at (m, 6), ｜ 6 = - 3M, ｜ M = - 2, ｜ − 3 = B6 = − 2K + B, the analytic expression of K = − 92B = − 3 ｜ line is y = − 92x − 3

Function f (x) = √ (3x ^ 2-2), if sequence an, A1 = 2, and an = f (a (n-1)), if BN = 3 ^ n / (an + an + 1), find SN of BN

an=f(a(n-1))=√[3(an-1)²-2
The two sides are square
an²-1=3(an-1²-1)
And A1 & # 178; - 1 = 3
So {an & # 178; - 1} is an equal ratio sequence with 3 as the first term and 3 as the common ratio
So an & # 178; - 1 = 3 ^ n
an=√(3^n +1)
BN = 3 ^ n / (an + an + 1) = 3 ^ n / {√ (3 ^ n + 1) + √ (3 ^ [n + 1) + 1]} use the square difference formula
=(3^n){√(3^n +1)-√(3^[n+1) +1]} /(-2 *3^n)
=-{√(3^n +1)-√(3^[n+1) +1]} /2
=-(an -an+1)/2
So Sn = - (a1-an + 1) / 2
={√[3^(n+1) +1] }/2 -1

How far is the earth from the moon?
384401 km or 384000 km.

384401 km is the average distance. The perigee of the moon is 363300 km, and the apogee is 405500 km

Known inequality ax squared - 3x + 2

ax²-3x+2

If the distance between the corresponding point of a number on the number axis and its opposite number on the number axis is 5 units of length, then the number is ()
A. 5 or - 5B. 52 or − 52C. 5 or − 52d. - 5 or 52

Let this number be a, then its opposite number is - A. according to the meaning of the title, we get | a - (- a) | = 5, 2A = ± 5, a = ± 52

Let f (x) be a function defined on R with period 2. In the interval [- 1,1], f (x) = {ax + 1 (1), - 1

∵ f (x) is a function defined on R with period 2, f (x) = ax + 1, - 1 ≤ x ＜ 0 BX + 2 x + 1, 0 ≤ x ≤ 1, ∵ f (3 / 2) = f (- 1 / 2) = 1-1 / 2 A, f (1 / 2) = B + 4 / 3; f (1 / 2) = f (3 / 2),