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Oracle Financials Cloud: General Ledger 2025 Implementation Professional

  • Exam Number/Code : 1z0-1054-25
  • Exam Name : Oracle Financials Cloud: General Ledger 2025 Implementation Professional
  • Questions and Answers : 213 Q&As
  • Update Time: 2019-01-10
  • Price: $ 99.00 $ 39.00

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NEW QUESTION: 1
A customer has two redundant fabrics with four switches per fabric1 and each fabric has two VSAN5.
What is the minimum number of out-of-band connections required by Fabric Manager to discover and manage the fabrics?
A. 0
B. 1
C. 2
D. 3
Answer: D

NEW QUESTION: 2
To improve response times for an OLAP application, a customer requires a SAN-attached storage
solution with microsecond latency and highest density per rack.
Which of the following supports these requirements?
A. Storwize V3700
B. EXP30
C. FlashSystem 900
D. Storwize V7000
Answer: B

NEW QUESTION: 3
Your company has an Exchange Server 2010 Service Pack 1 (SP1) organization named tailspintoys.com. You plan to implement a hybrid deployment of Office 365. You need to move 55 mailboxes from the Exchange on-premises environment to Office 365. Which cmdlet should you use?
A. New-MoveRequest
B. New-ProvisioningRequest
C. New-MigrationBatch
D. New-Mailbox
Answer: A

NEW QUESTION: 4
You administer a Microsoft SQL Server 2012 database named ContosoDb. Tables are defined as shown in the exhibit. (Click the Exhibit button.)

You need to display rows from the Orders table for the Customers row having the CustomerId value set to
1 in the following XML format.

Which Transact-SQL query should you use?
A. SELECT Name AS '@Name', Country AS '@Country', OrderId, OrderDate, Amount FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId WHERE Customers.CustomerId = 1 FOR XML PATH ('Customers')
B. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML RAW,ELEMENTS
C. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML AUTO
D. SELECT Name, Country, OrderId, OrderDate, Amount FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML AUTO
E. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML RAW
F. SELECT Name, Country, OrderId, OrderDate, Amount FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML AUTO, ELEMENTS
G. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FOR XML AUTO, ELEMENTS
H. SELECT Name AS 'Customers/Name', Country AS 'Customers/Country', OrderId, OrderDate, Amount FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId WHERE Customers.CustomerId = 1 FOR XML PATH ('Customers')
Answer: F
Explanation:
Explanation/Reference:
Explanation: