A technician has just finished installing a new NIC into a customer’s desktop. Which of thefollowing should the technician do NEXT?()A. Perform a full system backup.B. Follow up with the customer, verifying satisfaction.C. Ask their manager to contact the

题目
A technician has just finished installing a new NIC into a customer’s desktop. Which of thefollowing should the technician do NEXT?()

A. Perform a full system backup.

B. Follow up with the customer, verifying satisfaction.

C. Ask their manager to contact the customer.

D. Perform a complimentary virus scan.


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更多“A technician has just finished installing a new NIC into a customer’s desktop. Which of thefollowing should the technician do NEXT?() ”相关问题
  • 第1题:

    在循环双链表的p结点之后插入s结点的操作是______。

    A.p→next=s; p→next→prior=s; S→prior=p; S→next=p→next;

    B.s→next=p; s→next=p→next; p→next=s; p→next→prior=s;

    C.p→next=s; s→prior=p; p→next→prior=s; s→next=p→next;

    D.s→prior=p; s→next=p→next; p→next→prior=s; p→next=S;


    正确答案:D

  • 第2题:

    在单链表结点p之后插入结点s,正确的操作是()。

    A.p.next=s;s.next=p.next;

    B.s.next=p.next;p.next=s;

    C.p.next=s;p.next=s.next;

    D.p.next=s.next;p.next=s;


    s.next=p.next; p.next=s;

  • 第3题:

    16、在单链表结点p之后插入结点s,正确的操作是()。

    A.p.next=s;s.next=p.next;

    B.s.next=p.next;p.next=s;

    C.p.next=s;p.next=s.next;

    D.p.next=s.next;p.next=s;


    B

  • 第4题:

    在循环双链表的p节点之后插入s节点的操作是______。

    A.p→next=s; p→next→prior=s; s→prior=p; S→next=p→next;

    B.s→next=p; s→next=p→next; p→next=S; p→next→prior=s;

    C.p→next=s; s→prior=p; p→next→prior=s; s→next=p→next;

    D.s→prior=p; s→next=p→next; p→next→prior=s; p→next=s;


    正确答案:D

  • 第5题:

    在p所指结点后插入s所指结点的正确操作是:

    A.s->next =p+1; p->next=s;

    B.(*p).next=s; (*s).next=(*p).next;

    C.s->next=p->next; p->next=s->next;

    D.s->next=p->next; p->next=s;


    S->prior=p;s->next=p->next;p->next->prior=s;p->next=s;

  • 第6题:

    在一个单链表中,若删除s所指结点的直接后继结点,则执行()。

    A.p=s->next; s->next=p->next; free(p);

    B.p=s->next; free(p); s->next=s->next->next;

    C.s->next=s->next->next; free(s);

    D.s->next=s->next->next; p=s->next; free(p);


    B