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NEW QUESTION: 1
Bea can paint a house three times faster than Alice can paint a house. If, working together, it takes Alice and Bea 24 hours to paint a house, then how many hours would it take Bea to paint a house alone?
A. 0
B. 1
C. 2
D. 3
E. 4
Answer: D
Explanation:
Explanation/Reference:
Explanation:

NEW QUESTION: 2
展示を参照してください。

Link1は銅線接続で、Link2はファイバー接続です。ファイバポートは、すべての転送のプライマリポートである必要があります。 SW2でのshow spanning-treeコマンドの出力は、ファイバーポートがスパニングツリーによってブロックされていることを示しています。エンジニアがSW2のGO / 1でspanning-tree port-priority 32コマンドを入力しましたが、ポートはブロックされたままです。問題を解決するには、Link2に接続されているポートでどのコマンドを入力する必要がありますか？
A. SW2のスパニングツリーポートプライオリティ4を入力します。
B. SW1にスパニングツリーポートプライオリティ32を入力します。
C. SW1にスパニングツリーポートプライオリティ224を入力します。
D. SW2でスパニングツリーポートプライオリティ64を入力します。
Answer: B
Explanation:
Explanation
SW1 needs to block one of its ports to SW2 to avoid a bridging loop between the two switches.
Unfortunately, it blocked the fiber port Link2. But how does SW2 select its blocked port? Well, the answer is based on the BPDUs it receives from SW1. answer 'Enter spanning-tree port-priority 32 on SW1' BPDU is superior than another if it has:
1. answer 'Enter spanning-tree port-priority 32 on SW1' lower Root Bridge ID
2. answer 'Enter spanning-tree port-priority 32 on SW1' lower path cost to the Root
3. answer 'Enter spanning-tree port-priority 32 on SW1' lower Sending Bridge ID
4. answer 'Enter spanning-tree port-priority 32 on SW1' lower Sending Port ID These four parameters are examined in order. In this specific case, all the BPDUs sent by SW1 have the same Root Bridge ID, the same path cost to the Root and the same Sending Bridge ID.
The only parameter left to select the best one is the Sending Port ID (Port ID = port priority + port index). And the port index of Gi0/0 is lower than the port index of Gi0/1 so Link 1 has been chosen as the primary link.
Therefore we must change the port priority to change the primary link. The lower numerical value of port priority, the higher priority that port has. In other words, we must change the port-priority on Gi0/1 of SW1 (not on Gi0/1 of SW2) to a lower value than that of Gi0/0.

NEW QUESTION: 3
You have a database for a mission-critical web application. The database is stored on a SQL Server 2012 instance and is the only database on the instance. The application generates all T-SQL statements dynamically and does not use stored procedures.
You need to maximize the amount of memory available for data caching.
Which advanced server option should you modify?
A. Scan for Startup Procs
B. Allow Triggers to Fire Others
C. Optimize for Ad hoc Workloads
D. Enable Contained Databases
Answer: D
Explanation:
Explanation/Reference:
Explanation:
A contained database is a database that is isolated from other databases and from the instance of SQL Server that hosts the database. SQL Server helps user to isolate their database from the instance in 4 ways.
Much of the metadata that describes a database is maintained in the database. (In addition to, or instead of, maintaining metadata in the master database.) All metadata are defined using the same collation.
User authentication can be performed by the database, reducing the databases dependency on the logins of the instance of SQL Server.
The SQL Server environment (DMV's, XEvents, etc.) reports and can act upon containment information.
References: https://docs.microsoft.com/en-us/sql/relational-databases/databases/contained-databases

NEW QUESTION: 4
Sie verwenden eine Microsoft Azure SQL DataBase-Instanz. Die Instanz enthält eine Tabelle mit dem Namen Customers, die Spalten mit den Namen Id, Name und IsPriority enthält.
Sie müssen eine Ansicht mit dem Namen VwPriorityCustomers erstellen, die:
* gibt Zeilen von Customer mit dem Wert True in der Spalte IsPriority und zurück
* erlaubt nicht, dass Spalten in der zugrunde liegenden Tabelle geändert oder gelöscht werden.
Welche Transact-SQL-Anweisung müssen Sie ausführen?
A. CREATE VIEW VwPriorityCustomers
MIT SCHEMABINDING
WIE
SELECT ID, Name FROM dbo.Customers WHERE IsPriority = 1
B. CREATE VIEW VwPriorityCustomers
WITH VIEW_METADATA
WIE
SELECT ID, Name FROM dbo.Customers WHERE IsPriority = 1
C. CREATE VIEW VwPriorityCustomers
MIT VERSCHLÜSSELUNG
WIE
SELECT ID, Name FROM dbo.Customers WHERE IsPriority = 1
D. CREATE VIEW VwPriorityCustomers
WIE
SELECT ID, Name FROM dbo.Customers WHERE IsPriority = 1
MIT CHECK-OPTION
Answer: A
Explanation:
Erläuterung
SCHEMABINDING bindet die Ansicht an das Schema der zugrunde liegenden Tabelle (n). Wenn SCHEMABINDING angegeben ist, können die Basistabelle (n) nicht so geändert werden, dass die Sichtdefinition beeinflusst wird.
Verweise:
https://docs.microsoft.com/de-de/sql/t-sql/statements/create-view-transact-sql?view=sql-server-2017

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