People who are included in a group are () from others.A、fightingB、arguingC、stayingD、isolated

题目

People who are included in a group are () from others.

A、fighting

B、arguing

C、staying

D、isolated

相似考题

1.We need ______ here this evening.A、to stayB、stayC、stayingD、stayed

2.It is a bad habit to () others.A、make a fun ofB、make fun toC、make up forD、make fun of

3.The case() a lot of things,() a second－hand watch.A、contained; includingB、included; containingC、included; containedD、contained; included

4.对由SELECT—FROM—WHERE—GROUP—ORDER组成的SQL语句，其在被DBMS处理时，各子句的执序次序(43)。A．SELECT—FROM—GROUP—WHERE—ORDERB．FROM—SELECT—WHERE—GROUP—ORDERC．FROM—WHERE—GROUP—SELECT—ORDERD．SELECT—FROM—WHERE—GROUP—ORDER

更多“People who are included in a group are () from others.A、fightingB、arguingC、stayingD、isol ”相关问题

  • 第1题：

    Whichbuilt－incopiesvaluesfromalistitemintoarecordgroup?()

    A.RETRIEVE_LIST

    B.RETRIEVE_GROUP

    C.RETRIEVE_LIST_ITEM

    D.RETRIEVE_GROUP_FROM_LIST

    E.RETRIEVE_GROUP_FROM_LIST_ITEM


    参考答案：A

  • 第2题：

    Whichbuilt－incopiesvaluesfromarecordgroupintoalistitem?()

    A.POPULATE_LIST

    B.POPULATE_GROUP

    C.POPULATE_LIST_ITEM

    D.POPULATE_LIST_FROM_GROUP

    E.POPULATE_LIST_ITEM_FROM_GROUP


    参考答案：A

  • 第3题：

    对由SELECT--FROM—WHERE—GROUP--ORDER组成的SQL语句，其在被DBMS处理时，各子句的执行次序为()

    A.SELECT—FROM—GROUP—WHERE—ORDER

    B.FROM——SELECT--WHERE——GROUP——ORDER

    C.FROM——WHERE——GROUP——SELECT——ORDER

    D.SELECT——FROM——WHERE——GROUP——ORDER


    答案：C

  • 第4题：

    Yourapplicationrequiresthatyouprogrammaticallymanipulateanonqueryrecordgroup.Whichthreebuilt－inscanyouuse?()

    A.CREATE_GROUP

    B.ADD_GROUP_NOW

    C.POPULATE_GROUP

    D.ADD_GROUP_COLUMN

    E.CREATE_GROUP_FROM_QUERY


    参考答案：A, B, D

  • 第5题：

    若查询的结果仅限于平均分数超过80分的，则应（ ）。A．在GROUP子句的下一行加入：HAVING AVG（grade）＞80B．在GROUP子句的上一行加入：HAVING AVG（grade）＞80C．在FROM子句的下一行加入：HAVING AVG（grade）＞80D．在FROM子句的上一行加入：HAVING AVG（grade）＞80


    正确答案：A
    语法 SELECT column_name, aggregate_function(column_name) FROM table_name WHERE column_name operator value GROUP BY column_name

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