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NEW QUESTION: 1
A TM1 Operations Console shows a thread which is waiting to obtain a read lock on a cube.
Which thread is blocking it?
A. A thread which holds a read lock on the cube
B. A thread which is waiting to write to the cube
C. A thread which holds a write lock on the cube
D. A thread which holds an intent to write lock on the cube
Answer: C
Explanation:
Explanation/Reference:
Explanation:

NEW QUESTION: 2
A company wants to distribute iOS devices to only the employees in the Sales OU. They want to be able to do the following on these devices:
* Control password policies.
* Make corporate apps available to the users.
* Remotely wipe the device if it's lost or compromised
What two steps are required before configuring the device policies? (Choose two.)
A. Turn on Advanced Mobile Management for the domain.
B. Set up an Apple Push Certificate.
C. Deploy Apple Certificate to every device.
D. Set up Device Approvals.
E. Turn on Advanced Mobile Management for Sales OU
Answer: A,D

NEW QUESTION: 3
R1とR2が互いにOTPを実行した場合、R1はR2からEIGRPを介してネットワーク10.0.2.0/24について学習し、プレフィックス10.0.2.0/24をEIDプレフィックスとして扱い、アドバタイジングネクストホップを取得します。
このEIDプレフィックスのRLOCとして198.51.100.62。同様に、R2はR1からEIGRPを介してネットワーク10.0.1.0/24について学習し、プレフィックス10.0.1.0/24をEIDプレフィックスとして扱い、アドバタイジングネクストホップ192.0.2.31をこのEIDプレフィックスのRLOCとして使用します。両方のルーターで、この情報はLISPマッピングテーブルにデータを入力するために使用されます。 10.0.1.0/24からへのパケット
10.0.2.0/24はR1に到達し、通常のLISPと同じようにLISPマッピングテーブルを使用して、パケットをLISPでカプセル化し、198.51.100.62にトンネリングする必要があることを検出します。 LISPデータプレーンはOTPで再利用され、変更されません。ただし、ネイティブのLISPマッピングおよび解決メカニズムはEIGRPに置き換えられています。
参照：CCIE Routing and Switching V5.0 Official Cert Guide、Volume 1、Fifth Edition EIGRPでサポートされているが、OSPFではサポートされていない機能はどれですか。
A. 不等コストの負荷分散
B. 等コストの負荷分散
C. ルートの要約
D. ルートフィルタリング
Answer: A
Explanation:
EIGRP support unequal-cost load balancing via the "variance ..." while OSPF only supports equalcost load balancing.

NEW QUESTION: 4
文字「e」を含み、最後から2番目の位置に文字「a」を持つ、CUSTOMERSテーブルのすべての顧客の名を表示する必要があります。
どのクエリで必要な出力が得られますか？
A. SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, 'e')<>'' ANDSUBSTR(cust_first_name, -2, 1)='a';
B. SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, 'e')IS NOT NULL ANDSUBSTR(cust_first_name, 1, -2)='a';
C. SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, 'e')<>0 ANDSUBSTR(cust_first_name, -2, 1)='a';
D. SELECT cust_first_nameFROM customersWHERE INSTR(cust_first_name, 'e')<>0 ANDSUBSTR(cust_first_name, LENGTH(cust_first_name), -2)='a';
Answer: C

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