((X)_(1)+(X)_(2)+(X)_(3))/3 b.X1-|||-c. ((X)_(1)+(X)_(2))/2 d. (2(X)_(1)+(X)_(2)+(X)_(3))/4-|||-则无偏估计是 __ ,其中最有效的是 __

参考答案与解析:

相关试题

(x)_(1)+2(x)_(2)+2(x)_(3)geqslant 4-|||-(x)_(1)+4(x)_(2)leqslant 20-|||-(x)_(1)+8(x)_(2)+2(x)_(3)leq

(x)_(1)+2(x)_(2)+2(x)_(3)geqslant 4-|||-(x)_(1)+4(x)_(2)leqslant 20-|||-(x)_(1)+

  • 查看答案
  • 设x 1 2 4-|||-f(x)= 2 2 1 2-x-|||-1 2 一 x 2 4-|||-1 x x+3 x+6,证明:x 1 2 4-|||-f(x)= 2 2 1 2-x-|||-1 2

    设x 1 2 4-|||-f(x)= 2 2 1 2-x-|||-1 2 一 x 2 4-|||-1 x x+3 x+6,证明:x 1 2 4-|||-f(x)

  • 查看答案
  • dfrac (1)(x) C. -dfrac (1)({x)^2} D. dfrac (2)({x)^3}

    dfrac (1)(x) C. -dfrac (1)({x)^2} D. dfrac (2)({x)^3}

  • 查看答案
  • ) (x)_(1)+4(x)_(2)-2(x)_(3)+8(x)_(4)=2 -(x)_(1)+2(x)_(2)+3(x)_(3)+4(x)_(4)=1 (x)_(1)geqslant 0(j=1,2

    ) (x)_(1)+4(x)_(2)-2(x)_(3)+8(x)_(4)=2 -(x)_(1)+2(x)_(2)+3(x)_(3)+4(x)_(4)=1 (x)

  • 查看答案
  • 1.已知方程组 ) (x)_(1)+(x)_(2)+(x)_(3)+(x)_(4)=2 3(x)_(1)+2(x)_(2)+(x)_(3)+(x)_(4)=a (x)_(2)+2(x)_(3)+2(

    1.已知方程组 ) (x)_(1)+(x)_(2)+(x)_(3)+(x)_(4)=2 3(x)_(1)+2(x)_(2)+(x)_(3)+(x)_(4)=a

  • 查看答案
  • 线性方程组 ) (x)_(1)+2(x)_(2)-2(x)_(3)=1 2(x)_(1)+4(x)_(2)-4(x)_(3)=2 3(x)_(1)+6(x)_(2)-6(x)_(3)=3 .

    线性方程组 ) (x)_(1)+2(x)_(2)-2(x)_(3)=1 2(x)_(1)+4(x)_(2)-4(x)_(3)=2 3(x)_(1)+6(x)_

  • 查看答案
  • 解方程_(2)-3(x)_(3)+4(x)_(4)=-5-|||-x1- 2x3+324 =-4-|||-(x)_(1)+2(x)_(2)-5(x)_(4)=12-|||-4x1+3x2-5x3 =5

    解方程_(2)-3(x)_(3)+4(x)_(4)=-5-|||-x1- 2x3+324 =-4-|||-(x)_(1)+2(x)_(2)-5(x)_(4)=1

  • 查看答案
  • 2.解方程组 ) (x)_(1)+(x)_(2)+(x)_(3) (x)_(1)+(x)_(2)-(x)_(3)-(x)_(4)=1 5(x)_(1)+5(x)_(2)-3(x)_(3)-4(x)

    2.解方程组 ) (x)_(1)+(x)_(2)+(x)_(3) (x)_(1)+(x)_(2)-(x)_(3)-(x)_(4)=1 5(x)_(1)+5(

  • 查看答案
  • 求二次型 ((x)_(1),(x)_(2),(x)_(3))=4({x)_(2)}^2-3({x)_(3)}^2+4(x)_(1)(x)_(2)-4(x)_(1)(x)_(3)+8(x)_(2)(x)

    求二次型 ((x)_(1),(x)_(2),(x)_(3))=4({x)_(2)}^2-3({x)_(3)}^2+4(x)_(1)(x)_(2)-4(x)_(1

  • 查看答案
  • 用消元法解下列线性方程组:-|||- x1+3x2+5x3-4x4 =1, x1+3x2+2x3-2x4+x5=-1, x1-2x2+x3-x4-x5=3, x1-4x2+x3+x4-x5=3, x1

    用消元法解下列线性方程组:-|||- x1+3x2+5x3-4x4 =1, x1+3x2+2x3-2x4+x5=-1, x1-2x2+x3-x4-x5=3, x

  • 查看答案