• 2022-06-15
    已知σ(X)=1,σ(Y)=2,X,Y相互独立,则σ(3X-2Y+4)=( )
    A: 4
    B: 11
    C: 14
    D: 20
  • A

    内容

    • 0

      方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$

    • 1

      已知直线的一般方程\( \left\{ {\matrix{ {x - 2y - z + 4 = 0} \cr {5x + y - 2z + 8 = 0} \cr } } \right. \), 则其点向式方程为( ) A: \( { { x - 2} \over 2} = {y \over { - 3}} = { { z - 4} \over {11}} \) B: \( {x \over 5} = {y \over { - 3}} = { { z - 4} \over {11}} \) C: \( { { x - 2} \over 5} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \) D: \( { { x - 2} \over 2} = { { y + 1} \over { - 3}} = { { z - 4} \over {11}} \)

    • 2

      已知E(X)= E(Y)=0, D(X)=1,D(Y)=4, ρXY =1/2 , 若Z=aX+Y与Y独立,则a等于( ) A: 2 B: -2 C: 4 D: -4

    • 3

      设 X ~ N(3, 12),Y ~ N(2, 4),且 X,Y 独立,则 X − Y ~ N(1, 8) .

    • 4

      设二维随机变量(X,Y)的联合密度函数如下,则X与Y相互独立的有() A: (1),(2) B: (1),(2),(3) C: (1),(2),(4) D: (3),(4)