急用!由E(xy)=E(x)E(y)能否推出D(xy)=D(x)D(y)?
举一反三
- 设\(z = {e^u}\sin v,\;u = xy,\;v = x + y\)，则\( { { \partial z} \over {\partial y}}=\)( ) A: \(x{e^{xy}}\sin \left( {x + y} \right) + {e^{xy}}\cos \left( {x + y} \right)\) B: \(x{e^{xy}}\sin \left( {x + y} \right) \) C: \( {e^{xy}}\cos \left( {x + y} \right)\) D: \(x{e^{xy}}\sin \left( {x + y} \right) - {e^{xy}}\cos \left( {x + y} \right)\)
- 设\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\) A: \({e^{xy}}({x^2}y + {y^3} + 2x)\) B: \({e^{xy}}({x}y^2 + {y^3} + 2x)\) C: \({e^{xy}}({x}y + {y^3} + 2x)\) D: \({e^{xy}}({x^2}y + {y^2} + 2x)\)
- 设X与Y是两个随机变量,则下列各式正确的是( ).(10.0) A: E(XY)=E(X)E(Y) B: D(XY)=D(X)D(Y) C: E(X+Y)=E(X)+E(Y) D: D(XY)=D(X)+D(Y)
- 设\(z = u{e^v}\)，\(u = x + y\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\) A: \({e^{xy}}(1 + xy + {y^2})\) B: \({e^{xy}}(1 + xy + {y^3})\) C: \({e^{xy}}(x+ xy + {y^2})\) D: \({e^{xy}}(y+ xy + {y^2})\)
- 设\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，则\( { { \partial z} \over {\partial y}}=\)（ ）。 A: \({e^{xy}}({x}y^2 + {x^3} + 2y)\) B: \({e^{xy}}({x^2}y + {x^3} + 2y)\) C: \({e^{xy}}({x}y^2 + {x^3} + 2x)\) D: \({e^{xy}}({x}y+ {x^3} + 2y)\)