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NEW QUESTION: 1
DRAG DROP
A customer's email server is not sending emails. Arrange the troubleshooting steps in the order the technician should perform them:
Answer:
Explanation:
NEW QUESTION: 2
Which of the following is NOT a form of detective technical control?
A. Intrusion detection system
B. Audit trails
C. Access control software
D. Honeypot
Answer: C
Explanation:
Explanation/Reference:
Explanation:
Access control software is an example of a preventive/technical control, not a detective/technical control.
By combining preventive and detective controls, types with the administrative, technical (logical), and physical means of implementation, the following pairings are obtained:
Preventive/administrative
Preventive/technical
Preventive/physical
Detective/administrative
Detective/technical
Detective/physical
The detective/technical control measures are intended to reveal the violations of security policy using technical means. These measures include intrusion detection systems and automatically-generated violation reports from audit trail information. These reports can indicate variations from "normal" operation or detect known signatures of unauthorized access episodes.
A honeypot is a system designed with the purpose of being attacked so that the attack can be monitored and the attack techniques noted. This is another example of a detective technical control.
Incorrect Answers:
A: Audit trails are an example of a detective/technical control.
C: A honeypot is an example of a detective/technical control.
D: An intrusion detection system is an example of a detective/technical control.
References:
Krutz, Ronald L. and Russell Dean Vines, The CISSP and CAP Prep Guide: Mastering CISSP and CAP, Wiley Publishing, Indianapolis, 2007, pp. 48-50
NEW QUESTION: 3
Your customer has asked you to come in and verify the operation of routers R1 and R2 which are configured to use HSRP. They have questions about how these two devices will perform in the event of a device failure.
What is the virtual mac-address of HSRP group 1?
A. 0000.0c07.ac01
B. 0000.0c07.0010
C. 4000.0000.ac02
D. 0000.0c07.ac02
E. 4000.0000.ac01
F. 4000.0000.0010
Answer: F
NEW QUESTION: 4
Let N(.) denote the cumulative distribution function and suppose that X and Y are standard normally distributed and uncorrelated. Using the fact that N(1.96)=0.975, the probability that X 0 and Y 1.96 is approximately
A. 0.25%
B. 0.488%
C. 0.495%
D. 0.49%
Answer: B
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