(X)_(9)}(sqrt {{{Y)_(1)}^2+({Y)_(2)}^2+... +({Y)_(9)}^2}}-|||-服从 __ _分布,统计量 =dfrac (5sum _{i=1)^5({x)_(i)}^2}(9sum _{i=1)^5({x)_(i)}^2} 服从_ __ _分布-|||-i=1

+(X)_(m))}(sqrt {{{Y)_(1)}^2}+({Y)_(2)}^2+... +({Y)_(n)}^2}} 服从t分布,则 dfrac (m)(n

设函数 (x,y)=1-dfrac (cos sqrt {{x)^2+(y)^2}}(tan ({x)^2+(y)^2)} ,则当定设函数 (x,y)=1-df

(9)设 f(x,y)= dfrac (1)({({x)^2+(y)^2)}^2},1leqslant xleqslant 3, dfrac (sqrt {3)

X_(9)是来自总体X的样本，Y_(1),Y_(2),... Y_(9)是来自总体Y的样本，则统计量U=(X_(1)+...+X_(9))/(sqrt(Y_(1

证明:函数-|||-f(x,y)= ((x)^2+(y)^2)sin dfrac (1)(sqrt {{x)^2+(y)^2}}, ^2+(y)^2neq 0,

7.设 (x,y)=dfrac (1)(xy),r=sqrt ({x)^2+(y)^2} _(1)= (x,y)|(x,y)in {R)^2 dfrac (1)

(5)已知积分区域 = (x,y)||x|+|y|leqslant dfrac {pi )(2)} _(1)=iint sqrt ({x)^2+(y)^2}d

,X_(9) 与 Y_(1),Y_(2), ... ,Y_(9) 是分别来自动体X,Y的简单样本,统计量 W= (X_(1)+X_(2)+ ... +X_(9)

(6) arctan dfrac (y)(x)=ln sqrt ({x)^2+(y)^2}

(B) dfrac (1)(2)(X)^2+dfrac (1)(2)(Y)^2 服从x^2分布.-|||-(C) dfrac (1)(3)((X+Y))^2 服