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Vendor: HP

Exam Code: HPE2-B02 Dumps

Questions and Answers: 104

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NEW QUESTION: 1
Refer to the exhibit.

What is the effect of the given command?
A. It configures encryption for MD5 HMAC.
B. It merges authentication and encryption methods to protect traffic that matches an ACL.
C. It configures the network to use a different transform set between peers.
D. It configures authentication as AES 256.
Answer: B
Explanation:
A transform set is an acceptable combination of security protocols, algorithms and other settings to apply to IP Security protected traffic. During the IPSec security association negotiation, the peers agree to use a particular transform set when protecting a particular data flow.
Source:
http://www.cisco.com/c/en/us/td/docs/ios/12_2/security/command
Explanation/Reference/srfipsec.html#wp1017694 To define a transform set -- an acceptable combination of security protocols and algorithms -- use the crypto ipsec transform-set global configuration command.
ESP Encryption Transform
+ esp-aes 256: ESP with the 256-bit AES encryption algorithm.
ESP Authentication Transform
+ esp-md5-hmac: ESP with the MD5 (HMAC variant) authentication algorithm. (No longer recommended) Source:
http://www.cisco.com/c/en/us/td/docs/ios-xml/ios/security/a1/sec-a1-cr-book/sec-cr-c3.html#wp2590984165

NEW QUESTION: 2
There are several students in a class with similar special needs. Which of the following would be LEAST likely to be considered proper accommodations for their needs?
A. Read the directions and questions aloud to those students
B. Give the test over several short periods
C. Allow an aide to stay with and encourage the student
D. Allow the student to take the test with the student's textbook and notes
Answer: D

NEW QUESTION: 3
You have two lists of values to correlate.

Which query lists all names in colors1 and how many total matches are there in colors2?
A. SELECT colors1.name, count (colors2.name)
FROM JOIN colors2
on colors1.name = colors2.name
GROUP BY colors1.name;
B. SELECT colors1.name, count(colorse2.name)
FROM colors1.name = colors2.name
WHERE colors1.name =colors2.name
GROUP BY colors1.name;
C. SELECT colors1.name count (colors2.name)
FROM colors1
INNER JOIN colors2
on colors1.name =colors2.name
GROUP BY colors1.name;
D. SELECT colors1.name.count (colors2.name)
FROM colors1.colors2
WHERE colors1.name = (SELECT DISTINCT name FROM colors2
WHERE colors1.name=colors2.name)
GROUP BY colors1.name;
E. SELECT colors1.name, count (colors2.name)
FROM colors1
RIGHT JOIN colors1
on colors1.name =colors2.name
GROUP BY colors1.name;
Answer: A


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