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NEW QUESTION: 1
Which of the following statements are most accurate about the Split-For-Each and Split-Join shapes? (Choose Two)
A. Both shapes allow you to continue processing when ANY or ALL of the sub processes complete
B. Both shapes create separate Threads for sub-processes they create
C. Split-Join allows you to execute different sub-flows whereas Split-For-Each calls the same process on different pages
D. Split-For-Each allows you to execute different sub-flows whereas Split-Join calls the same process on different pages
E. Split-For-Each can only be used when iterating over a list of work objects
Answer: A,C
NEW QUESTION: 2
写真共有Webサイトの管理者は、毎週、前週にユーザーがアップロードしたすべてのファイルのアーカイブを受け取ります。これらのファイルアーカイブは、サイズが10TBにもなることがあります。法的な理由により、これらのアーカイブは、誰かがこれらのアーカイブを削除または変更する可能性がない状態で保存する必要があります。コンテンツを表示する必要がある場合がありますが、コンテンツの取得には3時間以上かかる場合があります。
管理者は週次アーカイブをどうするべきですか？
A. AWS CLIを使用してアーカイブをAmazon Glacierにアップロードし、Vault Lockを有効にします。
B. 暗号化されたAmazon EBSボリュームでLinux EC2インスタンスを作成し、このインスタンスの毎週のアーカイブファイルをコピーします。
C. AWSマネジメントコンソールを介してファイルをAmazon S3にアップロードし、ライフサイクルポリシーを適用してストレージクラスをAmazon Glacierに変更します。
D. ストレージクラスS3低頻度アクセスを使用して、S3バケットのファイル共有に接続されたファイルゲートウェイを作成します。ゲートウェイ経由でアーカイブをアップロードします。
Answer: C
NEW QUESTION: 3
Sie verwalten eine Microsoft SQL Server 2012-Datenbank mit dem Namen ContosoDb. Die Tabellen sind wie im Exponat angegeben definiert. (Klicken Sie auf die Schaltfläche "Ausstellen".)
Sie müssen Zeilen aus der Orders-Tabelle für die Customers-Zeile mit dem CustomerId-Wert 1 im folgenden XML-Format anzeigen.
Welche Transact-SQL-Abfrage sollten Sie verwenden?
A. SELECT Name, Land, OrderId, OrderDate, Betrag FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FÜR XML AUTO ELEMENTE
B. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FÜR XML AUTO
C. SELECT Name AS '@Name', Land AS '@Country', OrderId, OrderDate, Betrag FROM Orders INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId WHERE Customers.CustomerId = 1 FOR XML PATH ('Customers' )
D. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FÜR XML RAW
E. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FÜR XML RAW ELEMENTE
F. WÄHLEN Sie Name, Land, Bestell-ID, Bestelldatum, Betrag von Bestellungen
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FÜR XML AUTO
G. WÄHLEN Sie den Namen ALS "Kunden / Name", das Land ALS "Kunden / Land", die Bestellnummer, das Bestelldatum und den Auftragsbetrag INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId WHERE Customers.CustomerId = 1 FOR XML PATH (' Kunden')
H. SELECT OrderId, OrderDate, Amount, Name, Country FROM Orders
INNER JOIN Customers ON Orders.CustomerId = Customers.CustomerId
WHERE Customers.CustomerId = 1
FÜR XML AUTO ELEMENTE
Answer: A