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NEW QUESTION: 1
Forward Inc. has renamed several departments. You want to use a Bulk Task to reflect these changes.
How do Bulk Tasks differ from the scheduled task functionality that CA Identity Minder provides?
A. They do not permit multi-valued attributes.
B. They are never scheduled.
C. The initiator is JInitiator.
D. They provide population filters.
Answer: D

NEW QUESTION: 2
ユーザーがAuto Scalingを使用してWebアプリケーションを作成しました。ユーザーは定期的にアプリケーションを監視しており、木曜日と金曜日の午前8時から午後6時の間にトラフィックが最も多いことを確認しました。この場合、スケーリングを処理するための最良のソリューションは何ですか？
A. 木曜日の午前8時までにスケールアップし、金曜日の午後6時以降にスケールダウンするようにAuto Scalingをスケジュールします。
B. 新しいインスタンスを木曜日の午前8時までに手動で追加し、金曜日の午後6時までに終了します。
C. 毎日午前8時にスケールアップし、午後6時にスケールダウンするポリシーをスケジュールします。
D. 午前8時までにインスタンスを追加し、金曜日の午後6時までにインスタンスを削除するようにバッチプロセスを構成します。
Answer: A
Explanation:
説明
スケジュールに基づいたAuto Scalingにより、ユーザーは予測可能な負荷の変化に応じてアプリケーションをスケーリングできます。この場合、負荷は木曜日までに増加し、金曜日までに減少します。したがって、ユーザーは、スケジュールによるAuto Scalingスケールを使用して、Webアプリケーションの予測可能なトラフィックパターンに基づいてスケーリングアクティビティをセットアップできます。

NEW QUESTION: 3
Which server is the source for active directory lightweight directory service information during the installation of a replica server?
A. security server
B. vcenter server
C. domain controller
D. connection server
Answer: D

NEW QUESTION: 4
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 8 kHz x 4-bit code words
C. 2 x 4-bit code words x 8 kHz
D. 2 x 4 kHz x 8-bit code words
Answer: D
Explanation:
Explanation/Reference:
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) Reference: 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."