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NEW QUESTION: 1
組織のソーシャルメディアコンテンツの保存期間はどのように定義する必要がありますか？
A. 各ソーシャルメディアサービスの保持ポリシーによる
B. 使用可能なストレージスペースの量による
C. 最高情報責任者（CIO）による
D. 組織の記録保持ポリシーによる
Answer: D
NEW QUESTION: 2
How do AMP's device trajectory capabilities help address customers' breach remediation issues?
A. It analyzes the data from suspicious files to provide a new level of threat intelligence
B. It determines the scope and cause of an outbreak and tracks suspicious files
C. It tracks file behavior across the network to see which devices it enters and exits
D. It searches for potential threats based on identified activities and behaviors
Answer: C
NEW QUESTION: 3
Refer to the exhibit.
Given the output from the show ip eigrp topology command, which router is the feasible successor?
A. 10.1.1.1 (Serial0.1), from 10.1.1.1, Send flag is 0x0
Composite metric is (46763776/46251776), Route is External
Vector metric:
Minimum bandwidth is 56 Kbit
Total delay is 41000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
B. 10.1.0.3 (Serial0), from 10.1.0.3, Send flag is 0x0
Composite metric is (46866176/46354176), Route is Internal
Vector metric:
Minimum bandwidth is 56 Kbit
Total delay is 45000 microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
C. 10.0.0.2 (Serial0.1), from 10.0.0.2, Send flag is 0x0
Composite metric is (53973248/128256), Route is Internal
Vector Metric:
Minimum bandwidth is 48 Kbit
Total delay is 25000 Microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 1
D. 10.1.0.1 (Serial0), from 10.1.0.1, Send flag is 0x0
Composite metric is (46152000/41640000), Route is Internal
Vector metric:
Minimum bandwidth is 64 Kbit
Total delay is 45000 Microseconds
Reliability is 255/255
Load is 1/255
Minimum MTU is 1500
Hop count is 2
Answer: C
NEW QUESTION: 4
あなたは、データベースアプリケーションを開発するためにMicrosoft SQL Server2012使用します。次の要件を満たすオブジェクトを作成する必要があります。
入力変数をとります。
値のテーブルを返します。
ビュー内で参照することはでき
どのオブジェクトを使うべきでしょうか。
A. ストアドプロシージャ
B. User-definedデータ・タイプ
C. inline table-valued function
D. Scalar-valued機能
Answer: C
Explanation:
Incorrect answers:
Not B: A user-defined data type would not be able to take an input parameter.
Not C: A stored procedure cannot be used within a view.
Not D: A scalar-valued would only be able to return a single simple value, not a table.