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NEW QUESTION: 1
What percent is X of Y?
(1) Y is bigger than 2X by 54.
(2) X is smaller than 3Y by 72.
A. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
B. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.
C. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Answer: A
Explanation:
Explanation/Reference:
Explanation:
Statement (1) can be written as: Y = 2X + 54.
Statement (2) can be written as: X = 3Y - 72.
Combining both statements, we have two different equations containing X and Y and so we can solve and find the value of X and Y and calculate what is X percent of Y.
The correct answer is C.

NEW QUESTION: 2
Which Unisphere for VMAX task must be performed to allow Unisphere to collect performance data on a Symmetrix array?
A. Set up Real Time Traces
B. Create Dashboards
C. System Registration
D. Set up recurring BTP report
Answer: C

NEW QUESTION: 3
次のテーブルを含むデータベースがあります。

すべての顧客IDと各顧客が最後に注文した日付のリストを作成する必要があります。 顧客が注文をしていない場合は、1900年1月1日の日付を返す必要があります。列名はCustomerIDとLastOrderDateでなければなりません。
ソリューションを開発するためにどの4つのTransact-SQLセグメントを使用する必要がありますか？ 回答するには、適切なTransact-SQLセグメントをTransact-SQLセグメントのリストから回答領域に移動して正しい順序で配置します。

Answer:
Explanation:

Explanation

Box 1: SELECT..COALESCE...
The COALESCE function evaluates the arguments in order and returns the current value of the first expression that initially does not evaluate to NULL.
Box 2: ..LEFT OUTER JOIN..
The LEFT JOIN (LEFT OUTER JOIN) keyword returns all rows from the left table (table1), with the matching rows in the right table (table2). The result is NULL in the right side when there is no match. A customer might have no orders so the right table must be allowed have a NULL value.
Box 3: ON c.custid = o.custid
We JOIN on the custID column, which is available in both tables.
Box 4: GROUP BY c.custid
References:
https://technet.microsoft.com/en-us/library/ms189499(v=sql.110).aspx
http://www.w3schools.com/sql/sql_join_left.asp

NEW QUESTION: 4
Which of the following will show you a history of job errors that have occurred in jobs scheduled to run in an SQL Server 2012 installation?
A. Look at the sys.all_job_hist table.
B. You must use T-SQL to write a job-logging procedure to monitor jobs.
C. In Management Studio, navigate to SQL Server Agent -> Error Logs and look at the log for the corresponding date.
D. Open the c:\error\job.log file and search for the particular job and the errors you wish to view.
Answer: B