A customer has just received three enterprise systems that support a total of 48 LPARs. Which installation method expedite the process?()A.Network installation managerB.Boot from installation mediaC.A migration installD.Copy the system using the HMC

题目
A customer has just received three enterprise systems that support a total of 48 LPARs. Which installation method expedite the process?()

A.Network installation manager

B.Boot from installation media

C.A migration install

D.Copy the system using the HMC


相似考题
更多“A customer has just received three enterprise systems that support a total of 48 LPARs. Which installation method expedite the process?() ”相关问题
  • 第1题:

    Despite the wonderful acting and well-developed plot the _________ movie could not hold our attention.

    A) three-hours B) three-hour

    C) three-hours’ D) three-hour’s

     

     


    选B

     用破折号连接的作形容词不用复数.

  • 第2题:

    当需要删除某个序列seq_customer_id,可以使用如下()方式。

    A.DROP SEQ_CUSTOMER_ID

    B.DELETE SEQUENCE SEQ_CUSTOMER_ID

    C.DROP SEQUENCE SEQ_CUSTOMER_ID

    D.DELETE SEQ_CUSTOMER_ID


    参考答案:C

  • 第3题:

    若要实现total=1+2+3+4+5求和,以下程序段错误的是()

    A.int i=1,total=1; while(i<5) { total+=i; i+=1; }

    B.int i=1,total=0; while(i<=5) { total+=i; i+=1; }

    C.int i=0,total=0; while(i<5) { i+=1; total+=i; }

    D.int i=0,total=0; while(i<=5) { total+=i; i+=1; }


    A

  • 第4题:

    Acustomer’sdatacenterhasallPOWER4+systems,andtheyplantoaddthreenewp5-570systems.Theywillbepurchasinganew7310-CR2HMCtosupportthePOWER5systems.WhatisakeyrequirementthattheyneedtoplanforwiththenewHMC?()

    A.Adesk,asitnotrackmountable

    B.EthernetconnectionstothePOWER5systems

    C.SerialconnectionstothePOWER5systems

    D.USBconnectionsforthekeyboardandmouse


    参考答案:B

  • 第5题:

    The ice is not thick enough to bear the weight of a tank.

    A:suffer
    B:accept
    C:receive
    D:support

    答案:D
    解析:

  • 第6题:

    34、若要实现total=1+2+3+4+5求和,以下程序段错误的是()

    A.int i=1,total=1; while(i<5) { total+=i; i+=1; }

    B.int i=1,total=0; while(i<=5) { total+=i; i+=1; }

    C.int i=0,total=0; while(i<5) { i+=1; total+=i; }

    D.int i=0,total=0; while(i<=5) { total+=i; i+=1; }


    A