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NEW QUESTION: 1
Which two tasks must you perform on a device to ensure that an Etherchannel operate at Layer
3? (Choose two.)
A. configure EtherChannel directly on the interface
B. configure switchport on
C. configure switchport mode access
D. switchport mode trunk
E. configure no switchport on physical interface
Answer: A,E

NEW QUESTION: 2
What is the most effective way to construct and organize the resource pools?
A. create resource pool placeholders ad-hoc since resources can only be added to pools
B. define the resource pool construct to reflect the practical concerns of the organization (geographical, organizational or virtual) to reflect the customers cross-functional requirements
C. define several pools all managed by the same group of managers
D. create a single resource pool managed by several managers
Answer: B

NEW QUESTION: 3
Refer to the exhibit. Which two pieces of information in this Wireshark capture indicate that you are viewing EIGRP traffic? (Choose two.)
A. the Class Selector
B. the header checksum
C. the header length
D. the source address
E. the destination address
F. the protocol number
Answer: E,F
Explanation:
Explanation/Reference:
Explanation:
EIGRP uses protocol number 88, which shows as EIGRP in the capture. Also, we in the capture that the destination IP address is 224.0.0.10, which is the Enhanced Interior Gateway Routing Protocol (EIGRP) group address is used to send routing information to all EIGRP routers on a network segment.

NEW QUESTION: 4
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.255.255.255.254
B. 許可10.0.0.1.0.0.0.0
C. 許可10.0.0.1.0.0.0.254
D. 許可10.0.0.0.0.0.0.1
Answer: C
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.