Juniper JN0-649 dumps - in .pdf

JN0-649 pdf
  • Exam Code: JN0-649
  • Exam Name: Enterprise Routing and Switching, Professional (JNCIP-ENT)
  • Version: V17.95
  • Q & A: 400 Questions and Answers
  • PDF Price: $51.98

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  • Exam Code: JN0-649
  • Exam Name: Enterprise Routing and Switching, Professional (JNCIP-ENT)
  • Version: V17.95
  • Q & A: 400 Questions and Answers
  • PDF Version + PC Test Engine + Online Test Engine
  • Value Pack Total: $103.96  $66.98
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Juniper JN0-649 dumps - Testing Engine

JN0-649 Testing Engine
  • Exam Code: JN0-649
  • Exam Name: Enterprise Routing and Switching, Professional (JNCIP-ENT)
  • Version: V17.95
  • Q & A: 400 Questions and Answers
  • Software Price: $51.98
  • Testing Engine

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NEW QUESTION: 1
Which parameters are set to the same value when a pair of BIGElIP devices are synchronized.
A. profile definitions
B. system clocks
C. VLAN failElsafe settings
D. host names
E. MAC masquerade addresses
Answer: A

NEW QUESTION: 2
BBB is a publisher. As a consequence of BBB using the Internet to outsource its typesetting and studio activities it has been able to move its few remaining staff into much smaller and less expensive office accommodation.
Which activity in Porter's Value Chain has benefitted from BBB's use of information systems?
A. HR management
B. Sales and marketing
C. Outbound logistics
D. Infrastructure
Answer: D

NEW QUESTION: 3
You are a database developer for an application hosted on a Microsoft SQL Server 2012 server. The database contains two tables that have the following definitions:

Global customers place orders from several countries. You need to view the country from which each customer has placed the most orders.
Which Transact-SQL query do you use?
A. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM Customer c INNER JOIN
(SELECT CustomerID, ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY OrderAmount DESC) AS Rnk
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WHERE o.Rnk = 1
B. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM Customer c INNER JOIN
(SELECT CustomerID, ShippingCountry,
COUNT(OrderAmount) DESC) AS OrderAmount
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
ORDER BY OrderAmount DESC
C. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM
(SELECT c.CustomerID, c.CustomerName, o.ShippingCountry, RANK()
OVER (PARTITION BY CustomerID
ORDER BY COUNT(o.OrderAmount) ASC) AS Rnk
FROM Customer c
INNER JOIN Orders o
ON c.CustomerID = o.CustomerID
GROUP BY c.CustomerID, c.CustomerName, o.ShippingCountry) cs WHERE Rnk = 1
D. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM Customer c INNER JOIN
(SELECT CustomerID, ShippingCountry,
RANK() OVER (PARTITION BY CustomerID
ORDER BY COUNT(OrderAmount) DESC) AS Rnk
FROM Orders
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WHERE o.Rnk = 1
Answer: D
Explanation:
Explanation/Reference:
Explanation:
Use descending (DESC) ordering.
To order by the number of orders we use ORDER BY COUNT(OrderAmount).
Finally a WHERE close is needed: WHERE o.Rnk = 1
Incorrect Answers:
B: The ascending (ASC) sorting would produce the country from which each customer has placed the least orders.
C: We are interested in the number of the orders, not the amount of the orders. We should use ORDER BY COUNT(OrderAmount), not ORDER BY OrderAmount.
D: We are only interested in one single post, only the country from which each customer has placed the most orders. Need to use a WHERE statement (here Where o.Rnk =1 ).

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