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NEW QUESTION: 1
When using Clustered Data ONTAP 8.1 and later, what is the default for new aggregates?
A. traditional aggregates
B. 64-bit aggregates
C. 32-bit aggregates
D. 128-bit aggregates
E. Vserver aggregates
F. cluster aggregates
Answer: B
NEW QUESTION: 2
Azure Data Lake Storage Gen2アカウントへのアクセスを提供するのはお客様の責任です。
ユーザーアカウントにはストレージアカウントへの寄稿者アクセス権があり、アプリケーションIDアクセスキーがあります。
PolyBaseを使用してAzure SQLデータウェアハウスにデータをロードする予定です。
データウェアハウスをストレージアカウントに接続するようにPolyBaseを構成する必要があります。
どの3つのコンポーネントを順番に作成する必要がありますか？回答するには、適切なコンポーネントをコンポーネントのリストから回答エリアに移動し、正しい順序で配置します。
Answer:
Explanation:
Explanation
Step 1: a database scoped credential
To access your Data Lake Storage account, you will need to create a Database Master Key to encrypt your credential secret used in the next step. You then create a database scoped credential.
Step 2: an external data source
Create the external data source. Use the CREATE EXTERNAL DATA SOURCE command to store the location of the data. Provide the credential created in the previous step.
Step 3: an external file format
Configure data format: To import the data from Data Lake Storage, you need to specify the External File Format. This object defines how the files are written in Data Lake Storage.
References:
https://docs.microsoft.com/en-us/azure/sql-data-warehouse/sql-data-warehouse-load-from-azure-data-lake-store
NEW QUESTION: 3
You move a switch from a network where it performed Layer 2 switching only to a network where it performs Layer 2 switching and Layer 3 routing. You notice that the switch is showing high CPU usage and significantly degraded routing performance. Which command sequence do you use to troubleshoot and fix the problem?
A. #show sdm prefer
(config)#sdm prefer default
B. #show sdm prefer
(config)#sdm prefer routing
C. #show running-config
(config)#ip routing
D. #show sdm prefer
(config)#sdm prefer vlan
Answer: B
NEW QUESTION: 4
Which two statements describe the IP address 10.16.3.65/23? (Choose two.)
A. The network is not subnetted.
B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
C. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
D. The subnet address is 10.16.3.0 255.255.254.0.
E. The last valid host address in the subnet is 10.16.2.254 255.255.254.0
Answer: B,C
Explanation:
The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in 0, 2, 4, 6, etc., all the way to 254. The host
10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254
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