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NEW QUESTION: 1
You have a database that contains the tables shown in the exhibit. (Click the Exhibit button).

You need to create a query for a report. The query must meet the following requirements:
NOT use object delimiters.
Return the most recent orders first.
Use the first initial of the table as an alias.
Return the most recent order date for each customer.
Retrieve the last name of the person who placed the order.
Return the order date in a column named MostRecentOrderDate that appears as the last column in the report.
The solution must support the ANSI SQL-99 standard.
Which code segment should you use?
To answer, type the correct code in the answer area.
A. SELECT LastName, O.OrderDate AS MostRecentOrderDate FROM Customers AS C INNER JOIN Orders AS O ON CustomerID = O.CustomerID ORDER BY O.OrderDate DESC
B. select C.Lastname, P.MostRecentOrderDate from customers AS C inner join ( select customID, MostRecentOrderDate=max(orderDate) from orders group by customID
)P
on C.customerID=P.CustomerID
order by P.MostRecentOrderDate desc
C. SELECT c.CustomerID --optional c.LastName, max(o.OrderDate) 'MostRecentOrderDate' FROM Customer c LEFT OUTER JOIN Orders o ON o.CustomerID = c.CustomerID GROUP BY c.CustomerID, c.LastName ORDER BY 3 DESC
Answer: C
Explanation:
--Burgos -NO Command equal to Derek, but without subquery. I Think that is better put c. CustomerID to identifies duplicated LastName. I used "OUTER JOIN" to permit that customers that had "no orders" appears too. SELECT c.CustomerID --optional c.LastName, max(o.OrderDate) 'MostRecentOrderDate' FROM Customer c LEFT OUTER JOIN Orders o ON o.CustomerID = c.CustomerID GROUP BY c.CustomerID, c.LastName ORDER BY o.OrderDate DESC --\Burgos
According to the table spec, lastname is not unique key. So my answer is
select C.Lastname, P.MostRecentOrderDate from customers AS C inner join ( select customID, MostRecentOrderDate=max(orderDate) from orders group by customID )P on C.customerID=P.CustomerID order by P.MostRecentOrderDate desc
--Derek =================================== The previous one missed "Group By" and did not consider the case of duplicate last names. SELECT C.LastName, O.OrderDate AS MostRecentOrderDate FROM Customers AS C INNER JOIN Orders AS O ON C.CustomerID = O.CustomerID ORDER BY O.OrderDate DESC

NEW QUESTION: 2
従業員のジョーは、同僚にスマートフォンについてどれだけ知っているかを見せたいと考えています。 Joeは、サードパーティから企業のスマートフォンにインストールした無料の映画アプリケーションのデモを行います。 Joeの同僚は、アプリストアでアプリケーションを見つけることができませんでした。 Joeがアプリケーションをインストールできるのは次のうちどれですか？ （2つ選択）。
A. Tethering
B. Sideloading
C. Rooting/jailbreaking
D. Ad-hoc connections
E. Near-field communication
Answer: B,C

NEW QUESTION: 3
A request has been made to the SAN administrator to provision a new LUN. A LUN (LUN 6) is created, and the SAN administrator presents the LUN to an ESXi 5.x host, using the storage array SAN management software. The administrator looks at the Storage configuration for the ESXi host, and does not see the new LUN. Other LUNs from the same storage array are visible.
Which two conditions are most likely to cause the error described? (Choose two.)
A. The administrator has not performed a Rescan All on the ESXi host.
B. The multipathing policy has been set to Fixed for the LUN.
C. The LUN was not presented to the correct ESXi host.
D. Zoning on the Fibre Channel switches is not configured correctly.
Answer: A,C
Explanation:
The cause of the error is that the administrator didn't perform a rescan all of ESXi host. Whenever a new LUN is provisioned, a manual rescan all has to be performed to enable ESXi to discover new LUNs. The second possibility is that the new LUN was not presented to the ESXi host that was intended by the administrator.