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The（ ）Process Group consists of the processes used to complete the work defined in the project management plan to accomplish the project’s requirements. A．Planning B．Executing C．Monitoring and Controlling D．Close－out

题目

The（ ）Process Group consists of the processes used to complete the work defined in the project management plan to accomplish the project’s requirements. A．Planning B．Executing C．Monitoring and Controlling D．Close－out

相似考题

1.PPP中,第二个P是指()A.publicB.privateC.partnershipsD.process

2.GP的“4P”是指( )和Process。A、PeopleB、PhilosophyC、PerformanceD、Product

3.Which traversal method for a binary tree does the following Pascal code illustrate? Procedure traverse(p：pointer)； Begin Ifp<>NIL Then begin Traverse (p^．left)； Process(p)； Traverse(p^．right) End； End_A．pre_orderB．middle_orderC．noneD．last_order

4.The story _________ mainly of four parts.A.makesB.composesC.is consistedD.consists

参考答案和解析
正确答案：B
执行(Executing)过程组由为达成项目要求而在项目管理计划中定义的工作的过程组成。
更多“The( )Process Group consists of the processes used to complete the work defined in the p ”相关问题

  • 第1题：

    SPSS结果中P值即为()。

    A、sig

    B、Z

    C、meanrank

    D、N

    E、group


    答案：A

  • 第2题：

    Which traversal method for a binary tree does the following Pascal code illustrate? procedure traverse (p:pointer); begin if p<>nil then begin traverse(p ↑ .left); process(p); traverse(p ↑ .right); end end;

    A．preorder

    B．postorder

    C．reorder

    D．inorder


    正确答案：D

  • 第3题：

    2、若有以下说明和语句，则引用方式不正确的是？ struct worker{ int no; char *name; }work, *p=&work;

    A．work.no

    B．(*p).no

    C．p-＞no

    D．work-＞no


    work-＞no

  • 第4题：

    The nuclear family ____ two parents (mother and father) and their children.

    A: consists of

    B: consist of

    C: is consisted of

    D: consists


    参考答案：A

  • 第5题：

    若有以下说明和语句，则以下引用方式中不正确的是（）。 struct worker { int no; char *name; }work, *p=&work;

    A．work.no

    B．(*p).no

    C．p-＞no

    D．work-＞no


    D解析：?的优先级远高于'＆'和'*'。

  • 第6题：

    若有以下说明和语句，则引用方式不正确的是？ struct worker{ int no; char *name; }work, *p=&work;

    A．work.no

    B．(*p).no

    C．p-＞no

    D．work-＞no


    D

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