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NEW QUESTION: 1
One way to help mitigate personnel risks that may occur during the end of the project is to
______________
A. Prepare a staff release plan
B. Provide specific recognition to each team member who has worked on the project
C. Meet individually with each team member
D. Document the time each person is to work on the project in a resource calendar
Answer: A
Explanation:
Explanation/Reference:
Explanation:
The staff release plan determines the method and timing of releasing team members. Morale is improved if there are smooth transitions for the staff to upcoming projects. This staff release plan also helps to mitigate human resource risks that may occur. It is part of the staffing management plan, which is part of the human resource plan.

NEW QUESTION: 2
Note: This question is part of a series of questions that use the same set of answer choices. An answer choice may be correct for more than one question in the series.
You administer a SQL Server server that contains a database named SalesDb. SalesDb contains a schema named Customers that has a table named Regions. A user named UserA is a member of a role named Sales. UserA is granted the Select permission on the Regions table. The Sales role is granted the Select permission on the Customers schema.
You need to ensure that the following requirements are met:
The Sales role does not have the Select permission on the Customers schema.

UserA has the Select permission on the Regions table.

Which Transact-SQL statement should you use?
A. DENY SELECT ON Schema::Customers FROM UserA
B. REVOKE SELECT ON Object::Regions FROM Sales
C. DENY SELECT ON Object::Regions FROM UserA
D. EXEC sp_addrolemember 'Sales', 'UserA'
E. REVOKE SELECT ON Schema::Customers FROM Sales
F. DENY SELECT ON Schema::Customers FROM Sales
G. EXEC sp_droprolemember 'Sales', 'UserA'
H. REVOKE SELECT ON Schema::Customers FOR UserA
I. DENY SELECT ON Object::Regions FROM Sales
J. REVOKE SELECT ON Object::Regions FROM UserA
Answer: E
Explanation:
Explanation/Reference:
References:
http://msdn.microsoft.com/en-us/library/ms188369.aspx
http://msdn.microsoft.com/en-us/library/ms187750.aspx
http://msdn.microsoft.com/en-us/library/ff848791.aspx

NEW QUESTION: 3
A Trojan horse is a destructive program that masquerades as a benign application. The software initially appears to perform a desirable function for the user prior to installation and/or execution, but in addition to the expected function steals information or harms the system.

The challenge for an attacker is to send a convincing file attachment to the victim, which gets easily executed on the victim machine without raising any suspicion. Today's end users are quite knowledgeable about malwares and viruses. Instead of sending games and fun executables, Hackers today are quite successful in spreading the Trojans using Rogue security software.
What is Rogue security software?
A. A Fake AV program that claims to rid a computer of malware, but instead installs spyware or other malware onto the computer. This kind of software is known as rogue security software.
B. Rogue security software is based on social engineering technique in which the attackers lures victim to visit spear phishing websites
C. This software disables firewalls and establishes reverse connecting tunnel between the victim's machine and that of the attacker
D. A flash file extension to Firefox that gets automatically installed when a victim visits rogue software disabling websites
Answer: A

NEW QUESTION: 4
You are developing a Windows Communication Foundation (WCF) service. The following code defines and implements the service. (Line numbers are included for reference only.)

You need to ensure that two identical copies of the received message are created in the service.
Which code segment should you insert at line 12?
A. MessageBuffer buffer = message.
CreateBufferedCopy(8192);
Message msgCopy = buffer.CreateMessage();
Message returnMsg = buffer.CreateMessage();
B. MessageBuffer buffer = message.
CreateBufferedCopy(8192);
Message msgCopy = buffer.CreateMessage();
Message returnMsg = msgCopy;
C. Message msgCopy = message.
CreateBufferedCopy(8192) as Message;
Message returnMsg = message.
CreateBufferedCopy(8192) as Message;
D. Message msgCopy = message;
Message returnMsg = message;
Answer: A