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NEW QUESTION: 1
A network engineer designing an access layer that requires all uplinks to be active, furthermore, VLANs must span across the entire switch block. Which two design fulfill this requirement? (Choose two)
A. Layer 2 loop square
B. Layer 2 loop-free U
C. Layer 2 Flex Links
D. Layer 2 loop-free inverted U
Answer: A,D

NEW QUESTION: 2
Joey wants to configure NTP on R80 Security Management Server. He decided to do this via WebUI. What is the correct address to access the Web UI for Gaia platform via browser?
A. https://<Device_IP_Address>
B. https://<Device_IP_Address>:4434
C. https://<Device_IP_Address>:10000
D. https://<Device_IP_Address>:443
Answer: A
Explanation:
Explanation
Access to Web UI Gaia administration interface, initiate a connection from a browser to the default administration IP address: Logging in to the WebUI Logging in To log in to the WebUI:
https://<Gaia IP address>

NEW QUESTION: 3
Smiths Consultants Inc.の管理部門の従業員は、クライアントの1人との契約の有効期限が開始日より早いことを発見しました。このエラーを防ぐことができる対策は何ですか？
A. 技術的対策
B. 可用性測定
C. 組織尺度
D. 整合性測定
Answer: A

NEW QUESTION: 4
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. ShippingCountry) cs
WHERE Rnk = 1
B. 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,
C. 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
D. ShippingCountry) cs
WHERE Rnk = 1
E. SELECT CustomerID, CustomerName, ShippingCountry
FROM
(SELECT c.CustomerID, c.CustomerName,
F. 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,
G. 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
Answer: G

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