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NEW QUESTION: 1
A workload management (WLM) configuration can define a total concurrency level of _ for all user-defined queues Choose the correct answer:
A. 0
B. 1
C. 2
D. 3
Answer: A
Explanation:
A workload management (WLM) configuration can define a total concurrency level of 50 for all user-defined queues Reference:
http://docs.aws.amazon.com/redshift/latest/mgmt/amazon-redshift-limits.html

NEW QUESTION: 2

A. Option B
B. Option A
C. Option D
D. Option C
Answer: B

NEW QUESTION: 3
For security reasons, a cloud service that can be accessed from anywhere would make BEST use of:
A. single sign-on.
B. data locality
C. multifactor authentication.
D. replication.
Answer: C

NEW QUESTION: 4
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 retrieve the students who scored the highest marks for each subject along with the marks.
Which Transact-SQL query should you use?
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, 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
C. SELECT StudentCode as Code, RANK() OVER(ORDER BY AVG(Marks) DESC) AS Value FROM StudentMarks GROUP BY StudentCode
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 StudentCode 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 SubjectCode ORDER BY Marks ASC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
G. SELECT StudentCode as Code, DENSE_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: D
Explanation:
Explanation/Reference:
Explanation: