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问题 单项选择题

A模型=β0+β1X1i+β2X2i+μi的最小二乘回归结果显示，样本可决系数R2为0.92，样本容量为30，总离差平方和为500，则估计的标准误差为（）。

A.1.217

B.1.482

C.4.152

D.5.214

答案

参考答案：A

填空题

五十万零四十写作______省略万后面的尾数约是______万．
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[说明]
下面程序定义了一个类Shape及其成员函数，然后从类Shape派生出类Point、类Circle、类Cylinder，以及它们的成员函数。
[Java程序]
(1) Shape
public abstract void printShapeName(); //纯虚函数。
public double area()return 0.0;
public double volumn()return 0.0;

class Point extends Shape
private double x.y;
public Point (double a, double b)x=a; y=b;
public void setPoint (double a,double b)x=a;y=b;
final public double getX() return x;
final public double getY()return y;
public void printShapeName()
(2) ;


class Circle extends Point
private double radius;
public Circle(double r, double a, double b)
(3) ;
radius=r＞0 r:0;

public void setRadius (double r)
radius=r＞0 r:0;

public double getRadius () return radius;
public double area()return 3.14259 * radius * radius;
public void printshapeName ()
System.out.print ("circle: ") ;
;
;
class Cylinder (4)
private double height;
public Cylinder(double h, double r, double x, double y)
super (r, x, y) ;
height = h＞0 h:0;

public void setHeight (double h)
height = h＞0 h:0;

public double area () //计算表面积
return (5) ;

public double volumn ()
double r=super.getRadius () ;
return 3.14159 * r * r * height;

public void printShapeName ()
System. out .print ("cylinder: ") ;
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