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NEW QUESTION: 1
What is a task of the Smart Event Server?
A. Display the received events.
B. Assign a severity level to an event.
C. Analyze each IPS log entry as it enters the Log server.
D. Forward what is known as an event to the Smart Event Server.
Answer: B

NEW QUESTION: 2
An engineer needs to perform a backup of user roles and locales from Cisco UCS Manager to replicate the setting to a different fabric interconnect. The engineer wants to review the file before importing it to a target fabric interconnect. Which backup type must be selected to meet these requirements?
A. full state
B. system configuration
C. logical configuration
D. all configuration
Answer: B
Explanation:
Explanation
https://www.cisco.com/c/en/us/td/docs/unified_computing/ucs/ucs-manager/GUI-User-Guides/Admin-Managem

NEW QUESTION: 3
Which two of these are characteristics of the 802.1Q protocol? (Choose two.)
A. It is used exclusively for tagging VLAN frames and does not address network reconvergence following switched network topology changes.
B. It modifies the 802.3 frame header, and thus requires that the FCS be recomputed.
C. It is a trunking protocol capable of carrying untagged frames.
D. It includes an 8-bit field which specifies the priority of a frame.
E. It is a Layer 2 messaging protocol which maintains VLAN configurations across networks.
Answer: B,C
Explanation:
Explanation/Reference:
Explanation:

NEW QUESTION: 4
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(colorse2. Name)
FROM colorse1 .name =colors2.name
WHERE colors1. Name =colors2.name
GROUP BY colors1.name,
B. SELECT colors1. Name count (colors2.name)
FROM colors1
INNER JOIN colors2
on colors1. Name =colors2. Name
GROUP BY colors1 .name;
C. SELECT colors1.name, count (colors2.name)
FROM JOIN colors2
on colors1 .name =colors2.name
GROUP BY colors1.name;
SELECT colors1.name, count (colors2.name)
FROM colors1
RIGHT JOIN colors1
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 colorse1.name,
Answer: C