填空题The members of set O are the integer solutions of the inequality 3x-4 ≤11, and the members of set P are the integer solutions of the inequality-4x+5<-7. What is one member of the intersection of O and P ?____

题目
填空题
The members of set O are the integer solutions of the inequality 3x-4 ≤11, and the members of set P are the integer solutions of the inequality-4x+5<-7. What is one member of the intersection of O and P ?____

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  • 第1题:

    How can team spirit be brought up within a team?

    A、Managers should order all member work together.

    B、Coach set strict rules for all the team members to obey.

    C、Team members should be persuaded to cooperate with each other.

    D、Team members should learn to show off personal ability.


    参考答案:C

  • 第2题:

    A Windows Communication Foundation (WCF) service is generating a separate namespace declaration for each body member of a message contract,even though all body members share the same namespace. You need to simplify the XML representation of your message contract so that the namespace is only declared once. What should you do?()

    A. Declare a wrapper namespace for the message contract by using the WrapperNamespace property of the MessageContract attribute

    B. Explicitly set the Namespace property of all the MessageBodyMember attrbutes to the same namespace.

    C. Declare all of the body members as properties of a DataContract class and use the class as the only body member of the message contract.

    D. Declare all of the body members as properties of a separate MessageContract class and use the class as the only body member of the message contract.


    参考答案:C

  • 第3题:

    有如下程序: Private Sub Command1_Click() Dim k As Integer,m As Integer Dim p As Integer k=4:m=1 p=PC(k,m):Print p; p=PC(k,m):Print p End Sub Private Function PC(a As Integer,b As Integer) Static m As Integer,i As Integer m=0:i=2 i=i + m + 1 m=i + a + b PC=m End Function 程序运行后,输出的结果为

    A.4 6

    B.6 6

    C.8 8

    D.10 12


    正确答案:C
    解析:在Sub过程中,程序段先定义了3个Integer型变量k,m,P,并给k赋给初值4,m的初值为1,然后调用事件过程PC,并将它的值赋给p;在事件过程PC中定义了两个形参,参数的传送是通过引用实参,即将k,m的地址作为a,b的地址;在PC中,将m,I定义为静态变量,所以第一次调用后的值仍然保留,但是m,I分别都有赋值语句,将它们的值变为0,2,所以返回值不变。

  • 第4题:

    有如下的程序: Private Sub Command1_Click() Dim k As Integer,m As Integer Dim P As Integer k=4:m=1 P=Fun(k,m):Print P; P=Fun(k,m):Print P End Sub Private Function Fun(a As Integer,b As Integer) Static m As Integer,i As Integer in=5:i=2 i=i+m+1 m=i+a+b Fun=m\2 End Function 单击命令按钮后,输出结果为

    A.2 2

    B.4 4

    C.5 5

    D.6 6


    正确答案:D
    解析:在Sub过程中,程序段先定义了3个Integer型变量k,m,p,并赋给k初值为4,m的初值为1,然后调用事件过程Fun,并将它的值赋给p;在事件过程Fun中定义了两个形参,参数的传送是通过引用实参,即将k,m的地址作为a,b的地址;在Fun中,将m,i定义为静态变量,所以第一次调用后的值仍然保留,但是m,i分别都有赋值语句,将它们的值变为5,2,所以返回值不变,都是6,6。

  • 第5题:

    以下程序运行后,单击命令按钮,窗体显示的结果是 ______。 Private Function p1(x As Integer,y As Integel,z As Integer) p1=2*x+y+3*z End Function Private Function p2(x As Integer,y As Integer,z As Integer) p2=p1(2,y,x)+x End Function Private SubCommandl_Click() Dim a As Integer Dim b As Integer Dim c As Integer a=2:b=3:c=4 Print p2(c,b,A)

    End Sub

    A.19

    B.21

    C.22

    D.23


    正确答案:D
    解析:当程序执行到Print p2(c,b,a)语句时,首先调用函数过程p2,将实参c、b、a,的值4、3、2按顺序传递给p2的形参x、y、z,使形参x、y、z的值分别是4、3、2。当执行到函数过程p2中的语句p2=p1(z,y,x)+x时,又去调用函数过程p1,并将实参x、y、x的值2、3、4按顺序传递给形参x、y、z,使函数过程p1的形参x、y、z的值分别是2、3、4。于是执行p1中的语句p1=2*x+y+3*x=2*2+3+3*4=4+3+12=19。即函数过程p2中p1(2,y,x)的值为19,因此p2=p1(2,y,x)+x=19+4=23。该值返回给Command1_Click事件过程中的p2(c,b,a),所以窗体显示的值是23。

  • 第6题:

    有如下的程序:Private Sub Command1_Click( ) Dim k As Integer,m As Integer Dim P As Integer k=4:m=1 P=Pid(k,m) :Print P; P=Pid(k,m) :Print PEnd SubPrivate Function Pid(a As Integer,b As Integer) Static m As Integer,i As Integer m=5:i=2 i=i + m + 1 m=i + a + b Pid=m\2End Function单击命令按钮后,输出结果为

    A.2 2

    B.4 4

    C.5 5

    D.6 6


    正确答案:D
    解析:在Sub过程中,程序段先定义了3个Integer型变量k,m,p,并给k赋给初值4,m的初值为1,然后调用事件过程Pid,并将它的值赋给p;在事件过程Pid中定义了两个形参,参数的传送是通过引用实参,即将k,m的地址作为a,b的地址;在Pid中,将m,I定义为静态变量,所以第一次调用后的值仍然保留,但是m,I分别都有赋值语句,将它们的值变为5,2,所以返回值不变都是66。

  • 第7题:

    One of the redo log members in your database is lost. You queried V$logfile for further details. What would be the member’s status?()

    • A、LOST
    • B、INVALID
    • C、EXPIRED
    • D、DELETED
    • E、CORRUPT
    • F、UNKNOWN
    • G、OBSOLETE

    正确答案:B

  • 第8题:

    A Windows Communication Foundation (WCF) service is generating a separate namespace declaration for each body member of a message contract,even though all body members share the same namespace. You need to simplify the XML representation of your message contract so that the namespace is only declared once. What should you do?()

    • A、Declare a wrapper namespace for the message contract by using the WrapperNamespace property of the MessageContract attribute
    • B、Explicitly set the Namespace property of all the MessageBodyMember attrbutes to the same namespace.
    • C、Declare all of the body members as properties of a DataContract class and use the class as the only body member of the message contract.
    • D、Declare all of the body members as properties of a separate MessageContract class and use the class as the only body member of the message contract.

    正确答案:C

  • 第9题:

    问答题
    If 4 percent of (p + q) is 8 and p is a positive integer, what is the greatest possible value of q ?

    正确答案: 199
    解析:
    这是一道数学题,(p + q)的4%是8,因此p+q=200,设p=1,1是最小的正整数,则可知q的最大值是199。

  • 第10题:

    单选题
    What is the smallest integer value of x that satisfies the inequality 4-3x < 11?
    A

    -3

    B

    -2

    C

    -1

    D

    0

    E

    1


    正确答案: A
    解析:
    Solution 1: If 4-3x < 11, then -3x < 7, so x > -7/3. Since -7/3 is between -2 and -3, the smallest integer value of x that satisfies this inequality is -2. Solution 2: Plug each of the answer choices for x, starting with A., into 4-3x < 11 until you find one that makes the inequality a true statement. Choice A. gives 4-3(-3) < 11, 13 < 11 which is not a true statement. Choice B. gives 4-3(-2) < 11, 10 < 11 which is true, so there is no need to continue.

  • 第11题:

    单选题
    Windows Communication Foundation (WCF) application uses a data contract that has several data members. You need the application to throw a Serialization Exception if any of the data members are not present when a serialized instance of the data contract is deserialized. What should you do?()
    A

     Add the Known Type attribute to the data contract.Set a default value in each of the data member declarations.

    B

     Add the Known Type attribute to the data contract.Set the Order property of each data member to unique integer value.

    C

     Set the Emit Default Value property of each data member to false.

    D

     Set the Is Required property of each data member to true.


    正确答案: D
    解析: 暂无解析

  • 第12题:

    填空题
    下列程序的运行结果是()。  Program main    implicit none     integer, target :: a=1       integer, pointer :: p        p=>a        p=3    write(*,*)a  end

    正确答案: 3
    解析: 暂无解析

  • 第13题:

    设程序中有如下数组定义和过程调用语句: Dim a(10)As Integer … Call D(a) 如下过程定义巾,正确的是( )。

    A.Private Sub p(a As Integer)

    B.Private Sub p(a( )As Integer)

    C.Private Sub p(a(10)As Integer)

    D.Private Sub p(a(n)As Integer)


    正确答案:B
    B。【解析】本题考查的是过程的调用,根据题干在定义中不许定义一个a数组,其中a数组不能定长,所以答案为B。

  • 第14题:

    单击命令按钮时,下列程序的执行结果为 Private Sub Command1_Click() Dim a As Integer,b As Integer,c As Integer a=2:b=3:C=4 Print P2(c,b,A)End Sub Private Function P1(x As Integer,y As Integer,z As Integer) P1=2 * X + y + 3 * z End Function Private Function P2(x As Integer,y As Integer,z As Integer) P2=P1(z,x,y) + X End Function

    A.21

    B.19

    C.17

    D.34


    正确答案:A
    解析:对于多个过程或函数依次调用和处理与简单过程调用处理一样,本题先调用函数P2,而函数P2又调用函数P1,结果返回的顺序是从P1到P2,P2计算后输出到调用的地方。

  • 第15题:

    在窗体上画一个命令按钮,然后编写如下事件过程:Private Sub Command1_Click( ) Dim m As Integer,n As Integer,p As Integer m=3:n=5:p=O Call Y(m,n,p) Print Str(p)End SubSub Y(ByVal i As Integer,ByVal j As Integer,k As Integer) k=i + jEnd Sub程序运行后,如果单击命令按钮,则在窗体上显示的内容是

    A.4

    B.6

    C.8

    D.10


    正确答案:C
    解析:此程序考查了函数的调用,函数过程有3个形参,功能是将第一个和第二个形参的和赋给第三个形参,其中前两个形参是按值传递,不会影响实参的数值,即在调用函数过程的过程中,不变化实参的数值;而第三个形参是按址传递,传递的是实参的地址,这样实参会随着过程中形参的变化而变化,即改变实参的数值。在主调过程中定义了三个整型变量,分别赋值3、5、0,调用过程后m、n的数值仍然是3、5,而p的值不再是0而变为8,即m、n的和,这也是被调函数的作用。

  • 第16题:

    设程序中有如下数组定义和过程调用语句:

    Dim a(10) as integer

    ……

    Call p(a)

    如下过程定义中,正确的是

    A)Private Sub p(a as integer)

    B)Private Sub p(a() as integer)

    C)Private Sub p(a(10) as integer)

    D)Private Sub p(a(n) as integer)


    正确答案:B
    【答案】:B
    【知识点】:以数组作为参数时形式参数的定义形式
    【解析】:形式参数为数组时,表达示为sub name( array() as integer[string......]),并不需要指明元素个数,但小括号不能省略。故本题选B。

  • 第17题:

    有如下的函数过程: Function Func2(a As Integer, b As Integer)'As Integer Static m As Integer, i As Integer m=0 i=2 a=i+m+1 b=i+a+b Func2=m End Function 在窗体上画一个命令按钮,然后编写如下事件过程: Private Sub Command1_Click() Dim p As Integer, k As Integer, m As Integer k=4 m=1 p=Fune2(k, m) Print k; m p = Func2(k, m) Print k; m End Sub 程序运行后,单击命令按钮,输出结果为( )。

    A.3 6<CR>3 6

    B.3 6<CR>3 11

    C.3 11<CR>3 6

    D.3 11<CR>3 11


    正确答案:B
    解析:本愿意在考查调用Func2时,参数是按地址传递的,所以两次调用Fune2时,实参k、m的值都改变了。程序运行过程如下:第一次调用Func2(k,m),k被行参a改变为3,m被行参改变为 6;第二次调用函数后,a为3,m为11。

  • 第18题:

    1. import java.util.*;  2. public class Example {  3. public static void main(String[] args) {  4. // insert code here  5. set.add(new integer(2));  6. set.add(new integer(l));  7. System.out.println(set);  8. }  9. }  Which code, inserted at line 4, guarantees that this program will output [1,2]? ()

    • A、 Set set = new TreeSet();
    • B、 Set set = new HashSet();
    • C、 Set set = new SortedSet();
    • D、 List set = new SortedList();
    • E、 Set set = new LinkedHashSet();

    正确答案:A

  • 第19题:

    Windows Communication Foundation (WCF) application uses a data contract that has several data members. You need the application to throw a Serialization Exception if any of the data members are not present when a serialized instance of the data contract is deserialized. What should you do?()

    • A、 Add the Known Type attribute to the data contract.Set a default value in each of the data member declarations.
    • B、 Add the Known Type attribute to the data contract.Set the Order property of each data member to unique integer value.
    • C、 Set the Emit Default Value property of each data member to false.
    • D、 Set the Is Required property of each data member to true.

    正确答案:D

  • 第20题:

    A Windows Communication Foundation (WCF) application uses a data contract that has several data members. You need the application to throw a SerializationException if any of the data members are not present when a serialized instance of the data contract is deserialized.What should you do?()

    • A、Add the KnownType attribute to the data contract. Set a default value in each of the data member declarations.
    • B、Add the KnownType attribute to the data contract. Set the Order property of each data member to unique integer value.
    • C、Set the EmitDefaultValue property of each data member to false.
    • D、Set the lsRequired property of each data member to true.

    正确答案:D

  • 第21题:

    单选题
    A Windows Communication Foundation (WCF) application uses a data contract that has several data members. You need the application to throw a SerializationException if any of the data members are not present when a serialized instance of the data contract is deserialized.What should you do?()
    A

    Add the KnownType attribute to the data contract. Set a default value in each of the data member declarations.

    B

    Add the KnownType attribute to the data contract. Set the Order property of each data member to unique integer value.

    C

    Set the EmitDefaultValue property of each data member to false.

    D

    Set the lsRequired property of each data member to true.


    正确答案: C
    解析: 暂无解析

  • 第22题:

    单选题
    One of the redo log members in your database is lost. You queried V$logfile for further details. What would be the member’s status?()
    A

    LOST

    B

    INVALID

    C

    EXPIRED

    D

    DELETED

    E

    CORRUPT

    F

    UNKNOWN

    G

    OBSOLETE


    正确答案: D
    解析: 暂无解析

  • 第23题:

    单选题
    1. import java.util.*;  2. public class Example {  3. public static void main(String[] args) {  4. // insert code here  5. set.add(new integer(2));  6. set.add(new integer(l));  7. System.out.println(set);  8. }  9. }  Which code, inserted at line 4, guarantees that this program will output [1,2]? ()
    A

     Set set = new TreeSet();

    B

     Set set = new HashSet();

    C

     Set set = new SortedSet();

    D

     List set = new SortedList();

    E

     Set set = new LinkedHashSet();


    正确答案: D
    解析: 暂无解析