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NEW QUESTION: 1
What is a key capability of HP LinkSite?
A. Automates consistent application of policies across all content based on meaning
B. Provides access to files and documents from anywhere on any device
C. Provides the ability to track outstanding actions
D. Implements an information governance strategy
Answer: D
Explanation:
Reference:
http://www8.hp.com/us/en/hp-news/press-release.html?id=1900260
NEW QUESTION: 2
An administrator needs to see a list of the system user's encrypted passwords. Which of the following Linux files does the administrator need to read?
A. /etc/passwd
B. /etc/skel
C. /etc/pw
D. /etc/shadow
Answer: A
NEW QUESTION: 3
WEP encrypted data algorithms use algorithms.
A. AES
B. RC4
C. Asymmetric encryption
D. CCMP
Answer: B
NEW QUESTION: 4
あなたは、あなたは、学生の年半ば成績を記録するためにStudentCode、SubjectCodeおよび成績カラムを持っているテーブルを作成します。 テーブルには様々な主題のために50の学生によって得られたマークがあります。あなたは、次の要件が満たされていることを確認する必要があります。
* 学生は彼らの平均点に基づいて評価されていなければなりません。
* 一人以上の学生が同じ平均点を持っている場合、同じランクはこれらの学生に与えられなければなりません。
* 同じランクが割り当てられる場合、連続するランクがスキップされなければなりません。
どのTransact-SQLクエリを使うべきでしょうか。
A. SELECT StudentCode as Code,
RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
B. SELECT Id, Name, Marks,
DENSE_RANK() OVER(ORDER BY Marks DESC) AS Rank
FROM StudentMarks
C. 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
D. 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
E. 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
F. SELECT StudentCode as Code,
NTILE(2) OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
G. 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
H. SELECT StudentCode as Code,
DENSE_RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
Answer: A
Explanation:
Reference:
http://msdn.microsoft.com/en-us/library/ms189798.aspx