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NEW QUESTION: 1
Queue managers QMGR1, QMGR2 and QMGR3 are members of cluster QC. An administrator created the following queues for an application: On QMGR1: DEFINEQALIAS(TEST.ALIAS) + TARGET(TEST.LOCAL) + DEFBIND(OPEN) + DEFPSIST(NO) On QMGR2 and QMGR3:
DEFINE QLOCAL(TEST.LOCAL) + CLUSTER(QC) + DEFBIND(OPEN) + CLWLPRTY(0) + CLWLRANK(0) +DEFPSIST(NO) The administrator verifies the setup by putting messages on TEST.ALIAS using the sample amqsput program but notices that all messages end up on the same queue manager (either QMGR2 or QMGR3). What could be the likely cause?
A. The queue TEST.ALIAS was not made part of the cluster.
B. The queue TEST.ALIAS was defined with the wrong DEFBIND option.
C. The CLWLWGHT attribute of the Cluster channels was not set to 50.
D. The CLWLPRTY was set to 0 instead of 50 on TEST.LOCAL in both the queue managers.
Answer: B
NEW QUESTION: 2
A document which describes how project scope will be managed and how scope changes will be integrated into the project is: (choose the best answer)
A. A project plan
B. A scope management plan
C. A scope statement
D. A risk analysis
Answer: B
NEW QUESTION: 3
注：この質問は、同じまたは類似の回答の選択肢を使用する一連の質問の一部です。回答の選択は、シリーズ内の複数の質問に対して正しい場合があります。各質問は、このシリーズの他の質問とは無関係です。
質問に記載されている情報と詳細は、その質問にのみ適用されます。
次の表に示すテーブルを含むDB1という名前のMicrosoft SQL Serverデータベースがあります。
TBL1とTBL2の間に外部キーの関係はありません。
TBL1とTBL2からのデータを使用してデータを返すクエリに必要な時間を最小限に抑える必要があります。
あなたは何をするべきか？
A. TBL1とTBL2にクラスタ化インデックスを作成します。
B. TBL1の既存のインデックスを削除してから、クラスタ化カラムストアインデックスを作成します。 TBL1にノンクラスタードカラムストアインデックスを作成します。 TBL2に非クラスタインデックスを作成します。
C. TBL1とTBL2の両方に一意の制約を作成します。TBL1とTBL2の列を組み合わせたパーティションビューを作成します。
D. TBL1の既存のインデックスを削除してから、クラスタ化カラムストアインデックスを作成します。 TBL1にノンクラスタードカラムストアインデックスを作成します。TBL2に変更を加えません。
E. tbl2にのみ非クラスタ化インデックスを作成します。
F. TBL1にクラスタ化インデックスを作成します。 TBL2にノンクラスタードインデックスを作成し、最も頻繁にクエリされる列をインクルード列として追加します。
Answer: F
NEW QUESTION: 4
You are administering a database stored in Automatic Storage management (ASM). The files are stored in the DATA disk group. You execute the following command:
SQL > ALTER DISKGROUP data ADD ALIAS '+data/prod/myfile.dbf' FOR '
+data/prod/myfile.dbf'
What is the result?
A. The file '+data.231.54769' remains in the same location and a synonym 'myfile.dbf' is created.
B. The file '+data.231.54769' is renamed as 'myfile.dbf', and copied to '+data/prod'.
C. The file '+data.231.54769' is physically relocated to '+data/prod' and renamed as 'myfile.dbf'.
D. The file 'myfile.dbf' is created in '+data/prod' and the reference to '+data.231.54769' in the data dictionary removed.
Answer: A
Explanation:
ADD ALIAS Use this clause to create an alias name for an Oracle ASM filename. The alias_name consists of the full directory path and the alias itself.