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NEW QUESTION: 1
D18912E1457D5D1DDCBD40AB3BF70D5D

Refer to the exhibit showing logs from the Expressway-C, a copy of the Expressway-E certificate, and the UC traversal zone configuration for the Expressway-C. An office administrator is deploying mobile and remote access and sees an issue with the UC traversal zone. The zone is showing "TLS negotiation failure". What is causing this issue?
A. In the UC Traversal Zone on the Expressway-C, the peer address is set to the IP of the Expressway-E, which is not a SAN entry in the Expressway-E certificate
B. The Expressway-C is missing the FQDN of Cisco UCM in the Common Name of its certificate
C. The Expressway-E does not have the FQDN of Cisco UCM listed as a SAN in its certificate
D. The Expressway-E certificate includes the Expressway-C FQDN as a SAN entry
Answer: C

NEW QUESTION: 2
A company has an Aruba Mobility Master (MM)-based solution that runs ArubaOS 8 and uses the default AirMatch and Client Macth settings. The ARM profile has been disabled for 5GHz radios. How are channel and transmit power settings managed for these radios?
A. with legacy ARM
B. with AirMatch
C. with legacy Client Match
D. with Rule Based Client Macth
Answer: B

NEW QUESTION: 3
保存要件の期間中、アーカイブのどの側面を定期的にテストする必要がありますか?
A. 可用性
B. 回復可能性
C. 移植性
D. 監査能力
Answer: B
Explanation:
In order for any archiving system to be deemed useful and compliant, regular tests must be performed to ensure the data can still be recovered and accessible, should it ever be needed, for the duration of the retention requirements.

NEW QUESTION: 4

A. 10.10.0.0/16 subnetted with mask 255.255.255.252
B. 10.10.1.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/18 subnetted with mask 255.255.255.252
D. 10.10.1.0/24 subnetted with mask 255.255.255.252
E. 10.10.0.0/23 subnetted with mask 255.255.255.252
Answer: E
Explanation:
Explanation
We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7
bits (because 2

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7 = 128).
The network used for point-to-point connection should be /30.
So our initial network should be 30 - 7 = 23.
So 10.10.0.0/23 is the correct answer.
You can understand it more clearly when writing it in binary form:
/23 = 1111 1111.1111 1110.0000 0000
/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)
Topic 4, IP Routing Technologies