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NEW QUESTION: 1
When the test-aaa command is used on the access control device to test the connectivity to the Radius server, the operation result is displayed successfully. However, the user cannot access the device normally. Possible reasons do not include the following options:
A. The service controller does not join AD domain in AD authentication scenario.
B. The user account or password is wrong configured.
C. In the wireless 802.1X scenario, no security profile is configured on the access control device.
D. The access layer switch does not enable EAP transparent transmission.
Answer: B

NEW QUESTION: 2
What is the preference value in the routing table used for?
A. It differentiates between multiple routes to a destination learned by different routing protocols.
B. It indicates the preferred egress interface.
C. It is used to differentiate local routes from remote routes.
D. It differentiates between multiple routes to a destination learned by the same protocol.
Answer: A

NEW QUESTION: 3
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10.0.0.0.0.0.0.1
C. 許可10.0.0.1.0.0.0.0
D. 許可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 .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.

NEW QUESTION: 4
従業員が解雇されたときにユーザーアカウントを管理するのに最も効果的なのは次のうちどれですか？
A. すべてのアクセスを削除するために終了した従業員ネットワークIDを無効にする。
B. 終了した従業員のアクセスを自動削除するプロセスを実装します。
C. 終了時に従業員ネットワークとシステムIDを削除する。
D. すべてのシステムおよびアプリケーションへの終了した従業員のユーザーアクセスを手動で削除します。
Answer: C