Simplified symbol + [+ (+ 3)]

+[+（+3）]=+[+3]=+3=3

The problem of symbol simplification
- [- (– 9)] denotes the opposite number of – (– 9), and – (– 9) denotes the opposite number of – 9, that is, 9
–〔–（–9）〕=–9
Although we know that the final result is – 9, what we don't understand is why there is such a step as – 9. Since the result is – 9, it should not be simplified, and – (– 9) represents the opposite number of – 9, so – [- (– 9)] = – 9 is not better?

9 refers to – (– 9) = 9

The sign of reduction - (- a)
() is a bracket

If a > 0, = a
If a = 0, = 0
If a < 0, = - A

Simplify - [+ (- 100)] symbol
Like - 100,

-[+(-100)]=-[-100]=100
I don't think I need to explain. It's very clear

It is known that ab vector = (2,4) is the vector a "B" after vector a = (5,1)=__

Since AB vector = (2,4) is translated according to vector a = (5,1), then point a is also translated according to vector a = (5,1), that is, vector AA '= a = (5,1)
Substituting a 'B = ab-aa' = (2-5,4-1) = (- 3,3)

Let the joint density function of random vector (x, y) be f (x, y) = {8xy, 0 ≤ x ≤ y ≤ 1,0
Let the joint density function of random vector (x, y) be f (x, y) = {8xy, 0 ≤ x ≤ y ≤ 1,
Try to find two edge density functions FX (x), FY (y) (2) of (1) (x, y) to judge the independence of X and y

F (x) = ∫ (- ∞, + ∞) f (x, y) dy = ∫ (x, 1) 8xydy = 4x (1-x & # 178;), 0 ≤ x ≤ 1, others 0
F (y) = ∫ (- ∞, + ∞) f (x, y) DX = ∫ (0, y) 8xydx = 4Y & # 179;, 0 ≤ x ≤ 1, others are 0
F (x, y) is not equal to f (x) f (y)

Let the probability density function of random vector (x, y) be f (x, y) = {1 / y * e ^ (- (x / y + y)) x > 0, Y > 0, others, and find P {x ≥ 1 ｜ y = 2)

=P (x > = 1, y = 2) / P (y = 2). For molecules, bring x = 2 into the function to get the function about X, and then integrate on the negative infinity to 1. The result of finding the molecule is a number. For the denominator, if it is a little troublesome, first ask for the edge density of F (x, y) to y. look up the formula and get a function of Y. then bring y = 2 in to calculate the denominator, Synthesis can get the answer. Here is not really the place to do the problem!

Let f (x, y) = (1 + XY) / 4 be the probability density function of two-dimensional random vector (x, y)

Var (x) = var (y) = 1 / 3

Let the joint density function of random vector (x, y) be p (x, y) = {AE ^ - (2x + y), (x > 0, Y > 0); 0, others}
Find: (1) constant a (2) P {x

Let 1 = double integral [0, positive infinity] or directly observe that P (x, y) can be decomposed into the product of independent functions of X and y, so x and y are independent (this may not be mentioned in some textbooks, but it is true), and the coefficients are exponential distribution of 1 and 2, so 1x2 = 2
Double integral, upper and lower limits are 2,0 and 1,2 respectively to get (1-e ^ - 4) (1-e ^ - 1), or directly find the distribution function of exponential distribution
From the additivity of exponential distribution, we know that z = x + y ~ exponential (3)

Given the density function f (x, y) = 1 / 3 (x + y) of two-dimensional random vector (x, y), find the covariance cov (x, y)

University problem, very interesting, first ask for grace, remember is to first distinguish what distribution, and then find the probability distribution f (x, y), and then find the expectation e (x, y), variance D (x,. Y), and then find what autocorrelation, cross-correlation, (one remember seems to be generally 0), and then according to the covariance formula, I don't know, man, this class is over, remember when reviewing

Simplified symbol + [+ (+ 3)]