作业帮设D:x*2 y*2
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 04:45:07
回答：设Z=-Y,于是D(Z)=D(-Y),D(X-Y)=D(X)+D(-Y)=D(X)+D(Z)=1+2=3.
积分区域是圆S=πf(x,y)=1/π,-√（2y-y²）再问：没问题了
D(X-2Y+4)=D(X)+4D(Y)=4+40=44
用公式Cov(aX+bY,cW+dZ)=acCov(X,W)+bcCov(Y,M)+adCov(X,z)+bdCov(Y,Z)把数字往里代就可以了~还有Cov(X,X)=D(X)
有公式的D(X+_Y)=DX+DY+_2cov(X,Y)既然X,Y独立,协方差必为0D(X-Y)=DX+DY=3
D(X+Y)=E[(X+Y)^2]-E^2(X+Y).因为X与Y相互独立,所以E(XY)=E(X)E(Y);故上式等于:E(X^2+2XY+Y^2)-[E^2(X)+E^2(y)]=E(X^2)+E(
D(X-2Y)=D(X)+D(2Y)=D(X)+4D(Y)=25
取L：x²+y²+4x-2y≤0===>(x+2)²+(y-1)²≤5∮L(x²-y)dx+(-y²+2x)dy=∫∫D[∂/&
E[（X+Y）^2]=D（X+y）+[E(x+y)]^2,D(X+y)=D(x)+D(y)=2.E(x+y)=E(x)+E(y)=0;所以E[（X+Y）^2]=2不对么?
dy=dx+dsiny=dx+cosydy即y'=dy/dx=1/(1-cosy)对x求导y''=-1/(1-cosy)²*(1-cosy)'=-siny*y'/(1-cosy)²
E[(X+Y)^2]=E[(X-1+Y-1+2)^2]=E(X-1)^2+E(Y-1)^2+4+2*E(X-1)(Y-1)+2*2*E(X-1)+2*2*E(Y-1)=D(X)+D(Y)+4+0+0+
设y=2arctan(y/x),求dy/dx,d²y/dx².设F(x,y)=y-2arctan(y/x)=0,则dy/dx=-(∂F/∂x)/(ͦ
设随机变量X与Y相互独立并D（X）=4D（y）=2则D（X+2Y）=多少4+4＝8
设x^2=a则y=sin(a^2)∴dy/d(x^2)=dy/da=dsin(a^2)/da=cos(a^2)*da^2/da=2acos(a^2),将a=x^2代入式中即可得dy/d(x^2)=2x
cov（X,Y)=E(XY)-E(X)E(Y),这是协方差公式,但是你问的问题好像有问题哦,请把等号前面的字加上再问：不好意思，，，，设X,Y为随机变量,D(X)=4,D(Y)=16,Cov(X,Y)
E{[XY-E(XY)]^2}=E(X^2Y^2)-E(XY)^2=E(X^2)*E(Y^2)-E(X)^2*E(Y)^2=[D(X)+E(X)^2][D(Y)+E(Y)^2]-E(X)^2*E(Y)
首先看被积函数的几何意义注意到x²+y²+z²=R²是球体,所以z=√(R²-x²-y²)就是上半个球体半径为R,在xoy面的投影
极坐标∫∫√（a²+x²+y²）dxdy=∫∫r√(a²+r²)drdθ=∫[0→2π]dθ∫[0→a]r√(a²+r²)dr=2
d²y/dx²=d(dy/dx)/dx=d(4x³-16x)/dx=12x²-16,故选C.