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NEW QUESTION: 1
既存のマルチテナントコンテナデータベース（CDB）にマルチテラバイト非CDBを接続しようとしています。
非CDBの特徴は次のようにします:
-Version: Oracle Database 11g Release 2 (11.2.0.2.0) 64-bit -Character set: AL32UTF8 -National character set: AL16UTF16 -O/S: Oracle Linux 6 64-bit
CDBの特徴は次のようにします:
-Version: Oracle Database 12c Release 1 64-bit -Character Set: AL32UTF8 -National character set: AL16UTF16 -O/S: Oracle Linux 6 64-bit
どの方法はCDBにこの非CDBを接続しながら、ダウンタイムを最小限に抑えるために使用する必要がありますか。
A. RMAN
B. DBMS_PDBパッケージ
C. データ・ポンプのフルエクスポート/インポート
D. トランスポータブルデータベース
E. トランスポータブル表
Answer: E
Explanation:
* Overview, example:
- Log into ncdb12c as sys
- Get the database in a consistent state by shutting it down cleanly.
- Open the database in read only mode
- Run DBMS_PDB.DESCRIBE to create an XML file describing the database.
- Shut down ncdb12c
- Connect to target CDB (CDB2)
- Check whether non-cdb (NCDB12c) can be plugged into CDB(CDB2)
- Plug-in Non-CDB (NCDB12c) as PDB(NCDB12c) into target CDB(CDB2).
- Access the PDB and run the noncdb_to_pdb.sql script.
- Open the new PDB in read/write mode.
* You can easily plug an Oracle Database 12c non-CDB into a CDB. Just create a PDB manifest file for the non-CDB, and then use the manifest file to create a cloned PDB in the CDB.
* Note that to plug in a non-CDB database into a CDB, the non-CDB database needs to be of version 12c as well. So existing 11g databases will need to be upgraded to 12c before they can be part of a 12c CDB.

NEW QUESTION: 2
Which quality management tool would be used to determine potential causes of a production problem?
A. control chart.
B. scatter diagram.
C. Ishikawa diagram.
D. histogram.
E. run chart.
Answer: C

NEW QUESTION: 3
While migrating from a Power System E880 to a Power System E980, which of the following is automatically provided on both systems?
A. PowerVM Standard Edition
B. IBM Cloud PowerVC Manager
C. IBM Cloud Private
D. PowerVM Enterprise Edition
Answer: B

NEW QUESTION: 4
Which of the following lambda definitions are correct? (Select two answers)
A. lambda x, y; x\\y - x%y
B. lambda x, y; (x, y)
C. lambda (x, y = x\\y x%y
D. lanbda x, y; return x\\y - x%y
Answer: A,B

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