Y2~N(3,4) y2^2服从什么分布
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![Y2~N(3,4) y2^2服从什么分布](/uploads/image/f/907299-27-9.jpg?t=Y2%7EN%283%2C4%29+y2%5E2%E6%9C%8D%E4%BB%8E%E4%BB%80%E4%B9%88%E5%88%86%E5%B8%83)
你把两个函数图像在一个xy坐标轴上画出来,图像相交的部分就是他们的交集了!图像的应用在学交集的时候很适用
原式=[2(2x+y)]2-3x2-y2+2-2y2=4(2x+y)2-3(x2+y2)+2∵2x+y=7,x2+y2=5∴原式=183.
x2-xy=-3①,2xy-y2=-8②,①×2+②×3得:2x2-2xy+6xy-3y2=-6-24=-30,则2x2+4xy-3y2=-30.
A、原式=-6x2-19xy-5y2;B、原式=2x2-9xy-7y2;C、原式=x2-16xy-10y2;D、原式=8x2-13xy-15y2.故选D.
φ(x)=[1/(根号2π)]e^[-(x^2)/2]故:f(x,y)=φ(x)*φ(y)=[1/(2π)]e^[-(x^2+y^2)/2].故:E((X^2+Y^2)^(1/2))=∫∫[(x^2+
5x2—3xy—4y2—11xy—7x2+2y2=-2x2-2y2-14xy=-2(x2+y2)-14xy=-14+28=14
x+y=7xy=27^2=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2*2x^2+y^2=45(1)y=7-xx(7-x)=2x^2-7x+2=0x=(7+41^(1/2))/2或x=(
再问:你确定吗?再答:我保证再答:亲,那一步不明白可以问再问:你省略了一步吧!再答:能看明白吗再问:no再答: 再答:是不是这部再问:那么第一步是怎么得来的呢?再答:第一步不是你
已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*
(x²+y²)²+(x²+y²)-6-6=0(x²+y²)²+(x²+y²)-12=0(x²
(4x-3y)2-16y2=[(4x-3y)+4y][(4x-3y-4y]=(4x+y)(4x-7y).
原式=(2x²-3xy+4y²)+(x²+2xy-3y²)=(2x²+x²)+(-3xy+2xy)+(4y²-3y²)=
(1)2A-B=2(3x2+3y2-5xy)-(2xy-3y2+4x2)=6x2+6y2-10xy-2xy+3y2-4x2=2x2+9y2-12xy;(2)当x=3,y=−13时,2A-B=2x2+9
7x2-3xy-2y2-11xy-5x2+4y2=2x²-14xy+2y²=2(x²+y²)-14xy=2×7-14×(-2)=14+28=42
∵x2+y2=7,xy=-2,∴7x2-3xy-2y2-11xy-5x2+4y2,=(7x2-5x2)+(4y2-2y2)-(3xy+11xy),=2x2+2y2-14xy,=2(x2+y2)-14x
服从正态分布的随机变量的线性组合仍然服从正态分布,所以样本均值(X-Y)服从N(0,36)分布,(注:X-Y服从N(u1-u2,(σ1^2)/n1+(σ2^2)/n2).剩下的就是求正态分布的概率问题
6x2-xy-15y2=(2x+3y)(3x-5y)=0,所以x=-3/2y或x=5/3y
很难,因为不能变成系数为整数甚至分数的因式积.
1.y1=y2:-x+2=3x+4=>4x=-2=>x=-1/2y1-1/2y1>y2:-x+2>3x+4=>xx=1y=1带入y=ax+7=>a=-6
2x²+xy-3y²+x+4y-1=2×x×x+x×y-3×y×y+x+4×y-1=x×2x+x×y-y×3y+x×1+y×4-1=x×(2x+y+1)-y×(3y+4)-1再问: