参考答案和解析
参考答案:B
更多“They()since last night. They are about to finish the work.A. are cleaning the systemB. ”相关问题
  • 第1题:

    PDM includes four types of dependencies or precedence relationships.( ),the initiation of the successor activity,depends upon the initiation ofthe predecessor activity. A.Start-to-Start B.Finish-to-Finish C.Start-to-Finish D.Finish-to—Start


    正确答案:A
    前导图法包括四种活动依赖或前导关系。开始——开始(S-S),后续活动的开始依赖于前置活动的启动。

  • 第2题:

    PDM includes four types of dependencies or precedence relationships: ...(18). The completion of the successor activity depends upon the initiation of the predecessor activity.

    A.Finish-to-Start

    B.Finish-to-Finish

    C.Start-to-Start

    D.Start-to-Finish


    正确答案:D
    解析:PDM包括四种活动依赖或前导关系:...开始一结束。后续活动的结束依赖于前置活动的开始。

  • 第3题:

    Precedence Diagramming Method(PDM) is a method used in activity sequencing. There are four types of dependencies or precedence relationships in PDM. The initiation of the successor activity depends upon the completion of the predecessor activity is called ( ).

    A.Finish-to-Start
    B.Finish-to-Finish
    C.Start-to-Start
    D.Start-to-Finish

    答案:A
    解析:
    前导图法是用于活动排序的技术。前导图法包括四种活动依赖关系。先行活动结束后续活动才能开展的关系是结束—开始关系。

  • 第4题:

    PDM includes four types of dependencies or precedence relationships…(73),the initiation of the successor activity,depends upon the initiation of the predecessor activity.

    A.Finish-to-Start

    B.Finish-to-Finish

    C.Start-to-Start

    D.Start-to-Finish


    正确答案:C

  • 第5题:

    ● Precedence Diagramming Method (PDM) is a method used in activity sequencing. There are four types of dependencies or precedence relationships in PDM. The initiation of the successor activity depends upon the completion of the predecessor activity is called __(75)__.

    (75)

    A.Finish-to-Start

    B.Finish-to-Finish

    C.Start-to-Start

    D.Start-to-Finish


    正确答案:A

  • 第6题:

    已知last指向单向简单链表的尾结点,将s所指结点加在表尾,正确的操作是____。

    A.s->next=s,last=s,last->next=NULL;

    B.last->next=s,s->next=NULL,last=s;

    C.s->next=NULL, last->next=s, s=last;

    D.s->next=last, last->next=NULL,last=s;


    B