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NEW QUESTION: 1
Which statement is true about a Scheduler-generated event?
A. it indicates state changes that occur within the Scheduler itself.
B. It can be generated when a file arrives on the file system.
C. it requires jobs to be defined with an event condition and a queue specification.
D. it is raised by an application that is consumed by the Scheduler to start a job.
Answer: C
Explanation:
Consuming Job State Events with your Application To consume job state events, your application must subscribe to the Scheduler event queue SYS.SCHEDULER$_EVENT_QUEUE. This queue is a secure queue and is owned by SYS. To create a subscription to this queue for a user, do the following:
1.Log in to the database as the SYS user or as a user with the MANAGE ANY QUEUE
privilege.
2.Subscribe to the queue using a new or existing agent.
3.Run the package procedure DBMS_AQADM.ENABLE_DB_ACCESS as follows:
DBMS_AQADM.ENABLE_DB_ACCESS(agent_name, db_username); where agent_name
references the agent that you used to subscribe to the events queue, and db_username is
the user for whom you want to create a subscription.
There is no need to grant dequeue privileges to the user. The dequeue privilege is granted
on the Scheduler event queue to PUBLIC.

NEW QUESTION: 2
Given:

What is the result?
A. Three
B. Compilation fails.
C. The program runs, but prints no output.
D. One
Answer: A
Explanation:
Explanation/Reference:
Explanation:
push
void push(E e)
Pushes an element onto the stack represented by this deque (in other words, at the head of this deque) if it
is possible to do so immediately without violating capacity restrictions, returning true upon success and
throwing an IllegalStateException if no space is currently available.
This method is equivalent to addFirst(E).
pop
E pop()
Pops an element from the stack represented by this deque. In other words, removes and returns the first
element of this deque.
This method is equivalent to removeFirst().
Returns:
the element at the front of this deque (which is the top of the stack represented by this deque)
Throws:
NoSuchElementException - if this deque is empty

NEW QUESTION: 3
エンジニアは、高速移行を有効にしてWLANを構成します一部のレガシークライアントはこのWLANに接続できません。
レガシークライアントが接続しながら、他のクライアントがOLTIに基づく高速移行を使用できるようにする機能はどれですか。
A. アダプティブR
B. 802.11V
C. DS経由
D. 802.11k
Answer: A
Explanation:
802.11r Fast Transition（FT）ローミングは、802.11 IEEE規格の修正版です。ローミングの新しいコンセプトです。新しいAPとの最初のハンドシェイクは、クライアントがターゲットAPにローミングする前に行われます。そのため、これは高速移行と呼ばれます。 802.11rは、2つのローミング方法を提供します。
+無線：このタイプのローミングでは、クライアントは、高速遷移（FT）認証アルゴリズムを使用したIEEE 802.11認証を使用して、ターゲットAPと直接通信します。
+ Over-the-DS（配信システム）：このタイプのローミングでは、クライアントは現在のAPを介してターゲットAPと通信します。クライアントとターゲットAP間の通信は、クライアントと現在のAP間のFTアクションフレームで伝送され、コントローラーを介して送信されます。
ただし、これらの方法はどちらもレガシークライアントを処理しません。
802.11kにより、11k対応クライアントは、ローミングの候補である既知のネイバーAPに関する情報を含むネイバーレポートを要求できます。
リファレンス：https://www.cisco.com/c/en/us/td/docs/wireless/controller/technotes/80211r-ft/b-
80211r-dg.html
IEEE 802.11vは、IEEE 802.11規格の修正版であり、ワイヤレスネットワーク管理の多数の拡張機能について説明しています。このような機能強化の1つは、ネットワークを利用した省電力機能です。これにより、クライアントはより長くスリープできるようになり、バッテリー寿命が向上します。
もう1つの拡張機能は、ネットワーク支援ローミングです。これにより、WLANは関連付けられたクライアントに要求を送信し、より適切なAPに関連付けるようクライアントに通知できます。これは、負荷分散と接続が不十分なクライアントの指示の両方に役立ちます。
参照：https://www.cisco.com/c/en/us/td/docs/wireless/controller/9800/config-guide/b_wl_16_10_cg/802-11v.pdf Cisco 802.11rは3つのモードをサポートしています。
+ピュアモード：802.11rクライアントのみが接続できるようにします
+混合モード：FTをサポートするクライアントとサポートしないクライアントの両方が接続できるようにします
+アダプティブモード：FT AKMをまったくアドバタイズしませんが、サポートされているクライアントが接続するときにFTを使用します。したがって、ここでは「アダプティブモード」が最良の答えです。

NEW QUESTION: 4
What are the main models of H3C WA1208E? (Multiple choice)
A. WA1208E-AGP
B. WA1208E-AG
C. WA1208E-DG
D. WA1208E-G
E. WA1208E-GP
Answer: A,B,C,D,E