Draw the plane region represented by inequality LXL + Lyl ≤ 1, and find its area

Draw the plane region represented by inequality LXL + Lyl ≤ 1, and find its area

The area is 2

The geometry represented by equation √ x-3 * (x + y + 1) = 0 is

X-3 = 0 or x + y + 1 = 0
So it's two straight lines

Seeking: the equation of geometric figure
Example: triangle, rectangle, square
Parallelogram row
Stairway
Irregular lines, etc

Triangle: S = bottom * height / 2
Rectangle: S = length * width
Square: S = side length * side length
Parallelogram line: S = bottom * height
Trapezoid: S = (upper bottom + lower bottom) * height / 2

Calculation of mathematical area equation
The upper bottom is 15 meters, the lower bottom is 25 meters, the height is x meters, and the area is 160 square meters. What's the height of this trapezoid? Use the equation to calculate, according to the first knowledge point in unit 1 of Volume I of grade 6, the faster, the more points, and the more points will be added tomorrow

(15+25)*X÷2=160
X=160*2÷40
X=8

In covariance, cov (x, y) = e (XY) - E (x) e (y), how to calculate e (XY) here, for example

It's the same as e (x). E (x) = sigma X / N, so e (XY) = sigma (x * y) / N. in fact, that's it
For example? X1 = 3, X2 = 4, X3 = 8, Y1 = 2, y2 = 5, Y3 = 5
E(XY)=(3*2+4*5+8*5)/3=66/3=22

On the calculation formula of covariance of binary discrete random variables cov (x, y) = e (XY) - E (x) e (y), how can e (ey) be calculated?
E (x), e (y), (x), D (y) are known

The method is as follows
Cov(X,Y)=Σ(i=1->n) [Xi-E(X)][Yi-E(Y)] / {Σ(i=1->n) [Xi-E(X)]^2[Yi-E(Y)]^2}^0.5

What is e (XY) in the covariance formula and how to calculate it?

The expectation of x times y
See the expected formula for the algorithm

Random variable x y is not independent, x y is discrete random variable, how to calculate e (XY)

First make clear the distribution column of XY, and then do it according to the mean value calculation formula of discrete random variables
When x and y are not independent, P (AB) = P (a) P (B / a)

Let X and y be random variables and cov (x, y) = 3, then cov (2x, 3Y)=

According to the nature of covariance
CoV (ax, by) = abcov (x, y), (a, B are constants)
eighteen

Let (x, y) be a two-dimensional random variable, and prove that cov (x, y) = e (XY) - exey

E[(X-E(X))(Y-E(Y))]=E[XY-XE(Y)-E(X)Y+E(X)E(Y)]
=E(XY)-E(X)E(Y)-E(X)E(Y)+E(X)E(Y)
=E(XY)-E(X)E(Y)