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NEW QUESTION: 1
企業構造には、1つの元帳と2つのビジネスユニットがあります。ビジネスユニット1は、調達から支払いまでのビジネス機能でのみ購買依頼の予算管理を有効にし、ビジネスユニット2は、調達から支払いまでのビジネス機能でのみ支払請求の予算管理を有効にします。正しい2つのステートメントはどれですか？ （2つ選択してください。）
A. 各ビジネスユニットの予算管理例外を使用して、元帳レベルで予算管理を定義します。
B. ビジネスユニット1の管理を定義する際に、購買、支払請求、および受領の管理を無効にします。
C. ビジネスユニット1の管理を定義する際に、購入時の管理を有効にし、購買依頼のみを含むように例外を定義します。
D. ビジネスユニット2の管理を定義する際に、購買依頼で管理を有効にし、請求のみを含むように例外を定義します
E. 元帳レベルで予算管理を定義し、ビジネスユニットでのみ予算管理を定義します
F. ビジネスユニット2の管理を定義して、購買、購買、および受領の管理を無効にします。
Answer: B,F

NEW QUESTION: 2
DRAG DROP


Answer:
Explanation:

Explanation:

http://technet.microsoft.com/en-us/library/bb727008.aspx
File and Folder Permissions
On NTFS volumes, you can set security permissions on files and folders. These permissions grant or deny access to the files and folders.
File and Folder Permissions:

NEW QUESTION: 3
Sie arbeiten für ein Unternehmen, das mobile Anwendungen entwickelt. Sie unterhalten einen Server, auf dem die Spieleraufzeichnungen ihren verschiedenen Spielen zugewiesen werden. Das Trackingsystem ist neu und in Entwicklung.
Die Anwendung verwendet Entity Framework, um eine Verbindung zu einer Azure-Datenbank herzustellen. Die Datenbank enthält eine Spielertabelle und eine Spieltabelle.
Beim Hinzufügen eines Spielers sollte der Code einen neuen Spielerdatensatz einfügen und eine Beziehung zwischen einem vorhandenen Spielerdatensatz und dem neuen Spielerdatensatz hinzufügen.
Die Anwendung ruft CreatePlayerWithGame mit der richtigen gameId und der playerId auf, um den Vorgang zu starten.
(Zeilennummern dienen nur als Referenz.)

Wählen Sie für jede der folgenden Anweisungen Ja aus, wenn die Anweisung wahr ist. Andernfalls wählen Sie Nein.
HINWEIS: Jede richtige Auswahl ist einen Punkt wert.

Answer:
Explanation:

Erläuterung

Viele-zu-viele-Beziehungen ohne Entitätsklasse zur Darstellung der Verknüpfungstabelle werden noch nicht unterstützt. Sie können jedoch eine Viele-zu-Viele-Beziehung darstellen, indem Sie eine Entitätsklasse für die Verknüpfungstabelle einfügen und zwei separate Eins-zu-Viele-Beziehungen zuordnen.
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{
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HasKey (t => new {t.PostId, t.TagId});
modelBuilder.Entity <PostTag> ()
HasOne (pt => pt.Post)
WithMany (p => p.PostTags)
HasForeignKey (pt => pt.PostId);
modelBuilder.Entity <PostTag> ()
HasOne (pt => pt.Tag)
WithMany (t => t.PostTags)
HasForeignKey (pt => pt.TagId);
}
}