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NEW QUESTION: 1
次のステージのどれが予測カテゴリ「ベストケース」と一致する必要がありますか？
A. クローズして勝った
B. 閉鎖および紛失
C. パイプラインの初期段階
D. 後期パイプラインステージ
E. 中間パイプライン
Answer: E

NEW QUESTION: 2
Consider the following list of software components:
1.DCLI
2.Management Server (MS)
3.ASM Instance
4.RDBMS instance
5.Restart Server (RS)
6.Cellcli
7.Cell Server (CELLSRV)
8.Diskmon
Identify the location where these software components may run in the standard Database machine deployment.
A. 3, 4 and 8 run on the database servers; 1, 2, 5, 7 and 8 run on the Exadata Storage servers.
B. 1, 3, 4 and 8 run on the database servers; 1, 2, 5, 7 and 8 run on the Exadata Storage servers.
C. 3, 4 and 8 run on the database servers; 1, 2, 5, 6 and 7 run on the Exadata Storage servers.
D. 3, 4 and 8 run on the database servers; 1, 2, 5, 6 and 7 run on the Exadata Storage servers.
E. 4 and 8 run on the database servers; 1, 2, 3, 5, 6 and 7 run on the Exadata Storage servers.
F. 1, 3 and 4 run on the database servers; 2, 5, 6, 7 and 8 run on the Exadata Storage servers.
Answer: E
Explanation:
* (not C or notF):Exalogic includes a tool called DCLI (Distributed Command Line Interface) that can be used to run the same commands on all or a subset of compute nodes in parallel.
*Cellcli(not E): on Exadata Storage Servers.
*RDBMS instance on database server.
*CELLSRVon Exadata Storage Servers.
*ASM on Exadata Storage Servers(not D)
After an Oracle ASM instance has been installed on a single-instance Oracle Database or in an
Oracle Real Application Clusters (Oracle RAC) environment, the final Oracle ASM configuration
can be performed.

NEW QUESTION: 3
Which of these is the equation used to derive a 64 Kbps bit rate?
A. 8 kHz x 8-bit code words
B. 2 x 4-bit code words x 8 kHz
C. 2 x 4 kHz x 8-bit code words
D. 2 x 8 kHz x 4-bit code words
Answer: C
Explanation:
While the human ear can sense sounds from 20 to 20, 000 Hz, and speech encompasses sounds from about 200 to 9000 Hz, the telephone channel was designed to operate at about 300 to 3400 Hz. This economical range carries enough fidelity to allow callers to identify the party at the far end and sense their mood. Nyquist decided to extend the digitization to 4000 Hz, to capture higher-frequency sounds that the telephone channel may deliver. Therefore, the highest frequency for voice is 4000 Hz. According to Nyquist theory, we must double the highest frequency, so 2x4kHz = 8kHz.
Each sample will be encoded into a 8-bit code. Therefore 8kHz x 8-bit code = 64 Kbps (notice about the unit Kbps: 8kHz = 8000 samples per second so 8000 x 8-bit = 64000 bit per second = 64 Kilobit per second = 64 Kbps)
Link: http://encyclopedia2.thefreedictionary.com/Nyquist+theorem
Note:
Nyquist theory:
"When sampling a signal (e.g., converting from an analog signal to digital), the sampling frequency must be greater than twice the bandwidth of the input signal in order to be able to reconstruct the original perfectly from the sampled version."

NEW QUESTION: 4
エンジニアは、キーキャッシュメカニズムを展開することにより、ワイヤレスインフラストラクチャで発生している再認証の遅延を高速化する必要があります。どのメカニズムを構成する必要がありますか？
A. PMF
B. GTKランダム化
C. FT
D. PEAP
Answer: C
Explanation: