总体X服从 N(1,2)，现有样本X1，X2. 则 E（(X1+X2)/2）=？D（(X1+X2)/2）=？

x?x3?x2-3x3?x3+4x4?x2-2x2?x2?x2．

求函数 f(x)=3*x1^2 + 2*x1*x2 + x2^2 − 4*x1 + 5*x2. 时,输入代码 >>fun = @(x)3*x(1)^2 + 2*x(1)*x(2) + x(2)^2 - 4*x(1) + 5*x(2); >>x0 = [1,1]; >>[x,fval] = fminunc(fun,x0); 其中fun的作用是:

求函数 f(x)=3*x1^2 + 2*x1*x2 + x2^2 − 4*x1 + 5*x2. 时,输入代码 >>fun = @(x)3*x(1)^2 + 2*x(1)*x(2) + x(2)^2 - 4*x(1) + 5*x(2); >>x0 = [1,1]; >>[x,fval] = fminunc(fun,x0); 到matlab上运行一下,得到的结果,x是:

设f(x)与g(x)在(-∞，+∞)上都有定义，且x=x1是f(x)的唯一间断点，x=x2．是g(x)的唯一间断点．则( )． A: 当x1=x2时，f(x)+g(x)必有唯一的间断点x=x1． B: 当x1≠x2时，f(x)+g(x)必有两个间断点x=x1与x=x2． C: 当x1=x2时，f(x)g(x)必有唯一间断点x=x1． D: 当x1≠x2时，f(x)g(x)必有两个间断点x=x1与x=x2．

函数y=sin(2x−5)x的导函数为y=2xcos(2x−5)−sin(2x−5)x2y=2xcos(2x−5)−sin(2x−5)x2．

设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\)，则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)

数学式 A: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sinx^2-Cos2x)) B: (Exp(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) C: (Exp(2*x)*Ln(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) D: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x)^2-Cos(x)^2))

与数学关系式[img=114x22]1803bce8722f322.png[/img]等价的C语言关系表达式是？ A: x < -2 && x > 2 B: x < -2 || x > 2 C: -2 < x < 2 D: !(-2 <= x <=2) E: !(-2 <=x && x <= 2) F: x < -2, x > 2

求微分方程[img=101x35]17da5f15503f795.png[/img] 的通解,实验命令为(). A: dsolve(Dy+2*x*y=x*exp(-x^2))ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 B: dsolve('Dy+2*x*y=x*exp(-x^2)','x')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 C: dsolve('Dy+2*x*y=x*exp(-x^2)')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2