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NEW QUESTION: 1
VMware vSphere 6.x管理者がアラームアクションを作成しています。
3つの構成可能なアクションとは何ですか？ （3つ選択してください。）
A. Send a notification email.
B. Run a command.
C. Shutdown Guest.
D. Migrate VM.
E. Restart Guest.
Answer: A,B,D
NEW QUESTION: 2
暗号化されたVMをニューヨークからロンドンに移行する必要があります。この特定のVMは、転送中に暗号化する必要がある機密情報を格納します。
長距離vMotionトラフィックが飛行中に暗号化されるようにするために何をする必要がありますか？
A. 暗号化は物理ネットワーク層で処理する必要があります。
B. 長距離vSphere vMotionは暗号化をサポートしていません。
C. vSphere vMotionは、暗号化された仮想マシンを移行するときに暗号化を使用します。
D. 交通機関オプションで暗号化を選択します。
Answer: C
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10.0.0.1.0.0.0.254
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10.0.0.0.255.255.255.254
Answer: A
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 0.0.15.255 172.23.31.255.
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 000 ) 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.