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NEW QUESTION: 1
What is the maximum number of custom auto-attendant applications that you can set up in Cisco Unity Express?
A. 0
B. 1
C. 2
D. 3
E. 4
F. 5
Answer: B

NEW QUESTION: 2
See the Exhibit and examine the structure and data in the INVOICE table: Exhibit:

Which two SQL statements would executes successfully? (Choose two.)
A. SELECT MAX(inv_date), MIN(cust_id) FROM invoice;
B. SELECT MAX(AVG(SYSDATE - inv_date)) FROM invoice;
C. SELECT (AVG(inv_date) FROM invoice;
D. SELECT AVG(inv_date - SYSDATE), AVG(inv_amt) FROM invoice;
Answer: A,D

NEW QUESTION: 3
You administer a Microsoft SQL Server 2012 server. One of the databases on the server
supports a highly active OLTP application.
Users report abnormally long wait times when they submit data into the application.
You need to identify which queries are taking longer than 1 second to run over an extended period of time.
What should you do?
A. Use the Job Activity monitor to review all processes that are actively running. Review the Job History to find out the duration of each step.
B. use SQL Profiler to trace all queries that are processing on the server. Filter queries that have a Duration value of more than 1,000.
C. Use sp_configure to set a value for blocked process threshold. Create an extended event session.
D. Run the sp_who command from a query window.
E. Run the DBCC TRACEON 1222 command from a query window and review the SQL Server event log.
Answer: E
Explanation:
Verified the SQL Profiler and DBCC answers as correct. However, while Profiler will show this information, the best practice with Profiler is to use it short-term. The question specifically states "over an extended period of time". That means Profiler wouldn't be the best tool for this scenario. Therefore, DBCC would be the best answer.
http://www.mssqltips.com/sqlservertip/2130/finding-sql-server-deadlocks-using-trace-flag1222/ http://msdn.microsoft.com/en-us/library/ms188396.aspx

NEW QUESTION: 4

A. 192.168.1.16/28
B. 192.168.1.40/29
C. 192.168.1.8/29
D. 192.168.1.48/29
E. 192.168.1.32/27
Answer: B
Explanation:
Explanation
248 mask uses 5 bits (1111 1000)
42 IP in binary is (0010 1010)
The base subnet therefore is the lowest binary value that can be written without changing the output of an AND operation of the subnet mask and IP...
1111 1000 AND
0010 1010 equals
0010 1000 - which is .40
/24 is standard class C mask.
Adding the 5 bits from the .248 mask gives /29


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