Free H12-723 Exam Braindumps (page: 1)

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Which of the following options are right 802. 1X The description of the access process is correct? (Multiple choice).

  1. Through the entire authentication process, the terminal passes EAP The message exchanges information with the server.
  2. Terminal and 802.1X Switch EAP Message interaction,802.1X Switch and server use Radius Message exchange information
  3. 802.1X Authentication does not require security policy checks.
  4. use MD5 The algorithm checks the information.

Answer(s): B,D



BY00 The products and textiles provided by the history solution program cover the entire terminal network\Application and management and other fields/include: Serialization BC Equipment, paperless network system network access support, VPN Gateway, terminal security customer ladder software, authentication system, mobile device management(MDN),move eSpace UC.

  1. right
  2. wrong

Answer(s): A



For the terminal Wi-Fi The order of the push, which of the following is correct?
1. Any Office Mobile office system push Wi-Fi Configuration
2. Any Office The mobile office department automatically applies for a certificate.
3. The administrator configures the enterprise Wi-Fi Push.
4. The terminal automatically connects to the enterprise Wi-Fi.

  1. 1-2-3-4
  2. 4-2-3-1
  3. 3-2-1-4
  4. 2-3-1-4

Answer(s): C



Which of the following options is right PKI The sequence description of the work process is correct?
1. PKI Entity direction CA ask CA Certificate. .
2. PKI Entity received CA After the certificate, install CA Certificate.
3. CA receive PKI Entity CA When requesting a certificate, add your own CA Certificate reply to PKI entity.
4. PKI Entity direction CA Send a certificate registration request message.
5. PKI When the entities communicate with each other, they need to obtain and install the local certificate of the opposite entity.
6. PKI Entity received CA The certificate information sent.
7. PKI After the entity installs the local certificate of the opposite entity,Verify the validity of the local certificate of the peer entity.
When the certificate is valid,PC The public key of the certificate is used for encrypted communication between entities.
8. CA receive PKI The entity's certificate registration request message.

  1. 1-3-5-4-2-6-7-8
  2. 1-3-5-6-7-4-8-2
  3. 1-3-2-7-6-4-5-8
  4. 1-3-2-4-8-6-5-7

Answer(s): D



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Post your Comments and Discuss Huawei H12-723 exam with other Community members:

CCNA_owner 3/14/2024 2:12:24 AM
@ Serge 10/17/2023 2:34:12 AM question: QUESTION BEGINING IS BELOW: too (the fixed part is in bold: 2001:db8:0:0000). We borrowed 6 bits so we have to start from two next 00 and add 1 bit to the 6 bits we borrowed for each subnet (6 bits we borrowed are underlined): First subnet: 0000 0000 (binary) = 00 (hexadecimal) Second subnet: 0000 0100 (binary) = 04 (hexadecimal) Third subnet: 0000 1000 (binary) = 08 (hexadecimal) Therefore the third subnet will be 2001:db8:0:0008::/62 or 2001:db8:0:8::/62 in short. Required commands: On Sw101 Sw101(config)#int e0/0 Sw101(config-if)#ipv6 address 2001:DB8:0:8::/62 eui-64 On Sw102 (same commands as Sw101) Sw102(config)#int e0/0 Sw102(config-if)#ipv6 address 2001:DB8:0:8::/62 eui-64 Save the configuration Sw101#, Sw102#copy running-config startup-config
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CCNA_owner 3/13/2024 11:40:21 PM
@ Serge 10/17/2023 2:34:12 AM question: I encounter this lab on exam and have no idea how to solve it: A company plans to deploy 4 new sites. The sites will utilize both IPv4 and IPv6 networks 1. Subnet 10.20.0.0/16 to meet the subnet requirements and maximize the number of hosts • Using the third subnet o Assign the first usable IP address to e0/0 on Sw101 o Assign the last usable IP address to e0/0 on Sw102 2. Subnet 2001:db8::/52 to meet the subnet requirements and maximize the number of hosts • Using the third subnet o Assign an IPv6 GUA using a unique 64-Bit interface identifier on e0/0 on SW101 o Assign an IPv6 GUA using a unique 64-Bit interface identifier on e0/0 on SW102 Any ideas how to approach it? Answer: 64 new sites = 26 -> borrow 6 bits means the first subnet is 2001:db8::/62 Three first octets (2001:db8:0:) have 16 * 3 = 48 bits so they are fixed. 56 – 48 = 8 bits so the next 2 hexadecimal numbers are fixed too (the fixed part is in bold: 20
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