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NEW QUESTION: 1
Which statement is true?
A. An IPv6 address is 128 bits long and is represented as decimal digits.
B. An IPv6 address is 64 bits long and is represented as hexadecimal characters.
C. An IPv6 address is 32 bits long and is represented as decimal digits.
D. An IPv6 address is 128 bits long and is represented as hexadecimal characters.
Answer: D
Explanation:
Explanation/Reference: http://www.cisco.com/en/US/technologies/tk648/tk872/technologies_white_paper0900aecd 8026003d.pdf
One of the key advantages IPv6 brings is the exponentially larger address space. The following will outline the basic address architecture of IPv6. 128-bit-long addresses Represented in hexadecimal format: Uses CIDR principles: prefix/prefix length x:x:x:x:x:x:x:x, where x is a 16-bit hex field The last 64 bits are used for the interface ID

NEW QUESTION: 2
Sie sind Datenbankentwickler für eine Anwendung, die auf einem Microsoft SQL Server 2012-Server gehostet wird.
Die Datenbank enthält zwei Tabellen mit folgenden Definitionen:

Weltweite Kunden bestellen aus mehreren Ländern.
Sie müssen das Land anzeigen, aus dem jeder Kunde die meisten Bestellungen aufgegeben hat.
Welche Transact-SQL-Abfrage verwenden Sie?
A. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
VOM KUNDEN c
INNER JOIN
(SELECT Kundennummer, Versandland,
RANK () ÜBER (PARTITION NACH CustomerID
ORDER BY COUNT (OrderAmount) DESC) AS Rnk
FROM Bestellungen
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WO o.Rnk = 1
B. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
VOM KUNDEN c
INNER JOIN
(SELECT Kundennummer, Versandland,
COUNT (OrderAmount) DESC) ALS OrderAmount
FROM Bestellungen
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
BESTELLEN NACH OrderAmount DESC
C. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
VOM KUNDEN c
INNER JOIN
(SELECT Kundennummer, Versandland,
RANK () ÜBER (PARTITION NACH CustomerID
BESTELLUNG NACH BESTELLBETRAG DESC) ALS Rnk
FROM Bestellungen
GROUP BY CustomerID, ShippingCountry) AS o
ON c.CustomerID = o.CustomerID
WO o.Rnk = 1
D. SELECT c.CustomerID, c.CustomerName, o.ShippingCountry
VON
(SELECT c.CustomerID, c.CustomerName, o.ShippingCountry,
RANK () ÜBER (PARTITION NACH CustomerID
ORDER BY COUNT (o.OrderAmount) ASC) AS Rnk
VOM KUNDEN c
INNER JOIN Bestellungen o
ON c.CustomerID = o.CustomerID
GROUP BY c.CustomerID, c.CustomerName, o.ShippingCountry) cs
WO Rnk = 1
Answer: A

NEW QUESTION: 3

A. DAI
B. port security
C. IP Source Guard
D. BPDU guard
Answer: A