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NEW QUESTION: 1
왼쪽의 다음 보안 공학 용어를 오른쪽의 BEST 정의로 끌어옵니다.

Answer:
Explanation:

설명

위험 - 사건의 잠재적 상황, 상황 또는 사건이 발생할 경우 발생할 수있는 악영향, 발생 가능성에 의해 위협받는 범위에 대한 척도.
보호 필요성 평가 - 조직 및 시스템 자산에 대한 기밀성, 무결성 및 가용성 요구 사항을 식별하고 악영향이나 특성을 손실, 수정, 저하, 중단, 손상 또는 사용할 수 없게 만드는 특성을 특성화하는 데 사용되는 방법입니다.
위협 평가 (Threat Assessment) - 시스템의 수명주기 전반에 걸쳐 예상되는 위험을 식별하고 특성화하는 데 사용되는 방법.
보안 위험 처리 - 실현 가능한 보안 위험 완화 옵션 및 계획을 식별하는 데 사용되는 방법입니다.

NEW QUESTION: 2
注：この質問は、同じシナリオを提示する一連の質問の一部です。シリーズの各質問には、記載された目標を達成する可能性のある独自のソリューションが含まれています。一部の質問セットには複数の正しい解決策がある場合もあれば、正しい解決策がない場合もあります。
このセクションの質問に回答すると、その質問に戻ることはできません。その結果、これらの質問はレビュー画面に表示されません。
ネットワークにActive Directoryドメインが含まれています。
Microsoft Azure Active Directory（Azure AD）テナントを展開します。
別の管理者は、Azure ADと同期するようにドメインを構成します。
組織単位（OU）の10個のユーザーアカウントがAzure ADに同期されていないことがわかりました。他のすべてのユーザーアカウントは正常に同期されました。
Azure AD Connect Healthを確認し、すべてのユーザーアカウントの同期が正常に完了したことを発見します。
10個のユーザーアカウントがAzure ADに同期されていることを確認する必要があります。
解決策：Azure AD Connectから、Azure ADの資格情報を変更します。
これは目標を達成していますか？
A. はい
B. いいえ
Answer: B
Explanation:
The question states that "all the user account synchronizations completed successfully". Therefore, the Azure AD credentials are configured correctly in Azure AD Connect. It is likely that the 10 user accounts are being excluded from the synchronization cycle by a filtering rule.
Reference:
https://docs.microsoft.com/en-us/azure/active-directory/hybrid/how-to-connect-sync-configure-filtering

NEW QUESTION: 3
You create a table that has the StudentCode, SubjectCode, and Marks columns to record mid-year marks for students. The table has marks obtained by 50 students for various subjects.
You need to ensure that the following requirements are met:
* Students must be ranked based on their average marks.
* If one or more students have the same average, the same rank must be given to these students.
* Consecutive ranks must be skipped when the same rank is assigned.
Which Transact-SQL query should you use?
A. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY StudentCode ORDER BY Marks ASC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
B. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY SubjectCode ORDER BY Marks ASC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
C. SELECT StudentCode as Code,
RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
D. SELECT StudentCode as Code,
NTILE(2) OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
E. SELECT StudentCode as Code,
DENSE_RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
F. SELECT Id, Name, Marks,
DENSE_RANK() OVER(ORDER BY Marks DESC) AS Rank
FROM StudentMarks
G. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANXO OVER(PARTITION BY StudentCode ORDER BY Marks DESC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
H. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY SubjectCode ORDER BY Marks DESC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
Answer: C
Explanation:
Reference: http://msdn.microsoft.com/en-us/library/ms189798.aspx

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