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Exam Code: JN0-637
Exam Name: Security, Professional (JNCIP-SEC)
Vendor: Juniper

60 Questions & Answers
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NEW QUESTION: 1
What are the six stages of the HP Services lifecycle?
A. Design and Plan, Integrate, Set-up, Manage, Maintain, Repair and Refresh
B. Discover and Design, Configure, Deploy, Optimize, Maintain, Dispose and Refresh
C. Plan and Optimize, Configure, Install, Maximize, Maintain, Repair and Decommission
D. Discover and Configure, Deploy, Secure, Maintain, Replace and Decommission
Answer: C

NEW QUESTION: 2
The stock/requirements list for a finished product contains only one sales order and one fixed planned order with the same quantity. The availability dates for both MRP elements are on working days. The availability date for the fixed planned order is set to a later date than the one for the sales order.
However, a corresponding rescheduling exception message does not appear.
What could be the reasons? (Choose two)
A. The fixed planned order is outside the rescheduling horizon.
B. Requirements planning has not been carried out.
C. The MRP elements are linked via fixed pegging.
D. The delta between the availability dates falls within the tolerance window defined in Customizing.
Answer: A,D

NEW QUESTION: 3
Which of the following industries would not use process costing?
A. House-building
B. Food processing
C. Brewing
D. Chemical
Answer: A

NEW QUESTION: 4
You are a database developer of a Microsoft SQL Server 2014 database. You are designing a table that will store Customer data from different sources. The table will include a column that contains the CustomerID from the source system and a column that contains the SourceID. A sample of this data is as shown in the following table.

You need to ensure that the table has no duplicate CustomerID within a SourceID. You also need to ensure that the data in the table is in the order of SourceID and then CustomerID. Which Transact- SQL statement should you use?
A. CREATE TABLE Customer(SourceID int NOT NULL,CustomerID int NOT NULL,CustomerName varchar(255) NOT NULL,CONSTRAINT PK_Customer PRIMARY KEY CLUSTERED(SourceID, CustomerID));
B. CREATE TABLE Customer(SourceID int NOT NULL IDENTITY,CustomerID int NOT NULL IDENTITY,CustomerName varchar(255) NOT NULL);
C. CREATE TABLE Customer(SourceID int NOT NULL,CustomerID int NOT NULL PRIMARY KEY CLUSTERED,CustomerName varchar(255) NOT NULL);
D. CREATE TABLE Customer(SourceID int NOT NULL PRIMARY KEY CLUSTERED,CustomerID int NOT NULL UNIQUE,CustomerName varchar(255) NOT NULL);
Answer: A


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