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Accurate JN0-363 Answers, Valid JN0-363 Vce Dumps | Study JN0-363 Center - FreeTorrent

JN0-363

Exam Code: JN0-363

Exam Name: Service Provider Routing and Switching, Specialist (JNCIS-SP)

Version: V22.75

Q & A: 580 Questions and Answers

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NEW QUESTION: 1
次の展示に示すAzureリソースの階層があります。

RG1、RG2、およびRG3はリソースグループです。
RG2には、VM1という名前の仮想マシンが含まれています。
次の表に示すユーザーに、役割ベースのアクセス制御(RBAC)の役割を割り当てます。

次の各ステートメントについて、ステートメントがtrueの場合は、[はい]を選択します。それ以外の場合は、[いいえ]を選択します。
注:正しい選択はそれぞれ1ポイントの価値があります。

Answer:
Explanation:
Explanation


NEW QUESTION: 2
XPath expressions are often used by: Select the correct answer.
A. XSLT and XQuery
B. XQuery and SOAP
C. XSLT and WSDL
D. XQuery only
Answer: A

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 COUNT(OrderAmount) DESC) AS Rnk FROM Orders GROUP BY CustomerID, ShippingCountry) AS o ON c.CustomerID = o.CustomerID Where o.Rnk = 1
B. SELECT CustomerID, CustomerName, ShippingCountry FROM (SELECT c.CustomerID, c.CustomerName,
C. ShippingCountry) cs WHERE Rnk = 1
D. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry FROM Customer c INNER JOIN (SELECT CustomerID, ShippingCountry, COUNT(OrderAmount) AS OrderAmount FROM Orders GROUP BY CustomerID, ShippingCountry) AS o ON c.CustomerID = o.CustomerID ORDER BY OrderAmount DESC
E. ShippingCountry, RANK() OVER (PARTITION BY c. CustomerID ORDER BY o. OrderAmount DESC) AS Rnk FROM Customer c INNER JOIN Orders o ON c.CustomerID = o.CustomerID GROUP BY c.CustomerID, c.CustomerName,
F. SELECT CustomerID, CustomerName, ShippingCountry FROM (SELECT c.CustomerID, c.CustomerName, o.ShippingCountry, RANK() OVER (PARTITION BY c.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,
G. ShippingCountry) cs WHERE Rnk = 1
Answer: D

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