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NEW QUESTION: 1
Refer to the exhibit.

Which statement is true?
A. R4 does not run level1 IS-IS
B. R4 is a DIS
C. R4 is the DIS for level 1 and level 2
D. R5 is a DIS
Answer: B

NEW QUESTION: 2
Your work at ABC.com includes the preparation of company documents using Microsoft Office Word 2010. You have finished typing a 100 page document. You want the title of the document and the page number to appear at the bottom of every page. What is the easiest way to achieve this?
A. You should consider creating a Watermark.
B. You should consider creating a footer.
C. You should consider creating a template.
D. You should consider creating a footnote.
Answer: B

NEW QUESTION: 3
Refer to the Exhibit.

An engineer is installing a new pair of routers in a redundant configuration. When checking on the standby status of each router the engineer notices that the routers are not functioning as expected. Which action will resolve the configuration error?
A. configure matching hold and delay timers
B. configure unique virtual IP addresses
C. configure matching key-strings
D. configure matching priority values
Answer: C
Explanation:
Explanation
From the output exhibit, we notice that the key-string of R1 is Cisco123! (letter C is in capital) while that of R2 is cisco123!. This causes a mismatch in the authentication so we have to fix their key-strings.
key-string [encryption-type] text-string: Configures the text string for the key. The text-string argument is alphanumeric, case-sensitive, and supports special characters.
Reference:
https://www.cisco.com/c/en/us/td/docs/switches/datacenter/nexus9000/sw/6-x/security/configuration/guide/b_Ci

NEW QUESTION: 4
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.0.255.255.255.254
B. Permit 10.0.0.1.0.0.0.0
C. Permit 10.0.0.0.0.0.0.1
D. Permit 10.0.0.1.0.0.0.254
Answer: D
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.