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NEW QUESTION: 1
What dimension is an example of a common structure dimension category?
Please choose the correct answer.
Response:
A. Job Family
B. Grade / Band
C. Employment Level
D. Location
Answer: D
NEW QUESTION: 2
プロジェクトマネージャーには、アクティビティの推定所要時間に関する次の情報があります。
-ほとんどの場合[tM]-15日
-悲観的[tP]-20日
-楽観的[tO]-10日
三角分布法によると、この活動の推定期間はどれくらいですか?
A. 10日
B. 15日
C. 12.5日
D. 5日
Answer: C
NEW QUESTION: 3
An airline maintains a fleet of 4-engine aircraft. Its maintenance records show that on the average an engine fails 3 times in 10,000 operating hours with normal preventive maintenance.
What is the Poisson distributed probability that 2 or more engines on an aircraft will fail during a typical flying period of 8 hours?
Response:
A. 0.0000034
B. 0.000029
C. 0.000034
D. 0.0000029
Answer: D
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,
NTILE(2) OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
B. SELECT StudentCode as Code,
DENSE_RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
C. 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
D. 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
E. 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
F. 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
G. SELECT StudentCode as Code,
RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
H. SELECT Id, Name, Marks,
DENSE_RANK() OVER(ORDER BY Marks DESC) AS Rank
FROM StudentMarks
Answer: G
Explanation:
Reference:
http://msdn.microsoft.com/en-us/library/ms189798.aspx