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NEW QUESTION: 1
あなたは会社のBlueStarプロジェクトのプロジェクトマネージャーです。あなたの会社は機能組織として構成されており、定性リスク分析プロセスに移る準備ができていることを機能マネージャーに報告します。このシナリオでのプロジェクトの定性的リスク分析の入力として何が必要ですか？
A. リスクレジスター、リスク管理計画、定性的リスク分析の出力、および関連する組織プロセス資産が必要です。
B. リスク登録、リスク管理計画、機能マネージャーからの許可、および関連する組織プロセス資産が必要です。
C. リスクレジスター、リスク管理計画、プロジェクトスコープステートメント、および関連する組織プロセス資産が必要です。
D. 機能的な構造では、プロジェクトマネージャーを通じて定性的なリスク分析は行われません。
Answer: C

NEW QUESTION: 2
Sie haben eine Azure-Webanwendung mit dem Namen App1, die über zwei Bereitstellungssteckplätze mit den Namen Production und Staging verfügt. Jeder Steckplatz verfügt über die in der folgenden Tabelle angegebenen eindeutigen Einstellungen.

Sie führen einen Slot-Swap durch.
Wie sind die Konfigurationen des Produktions-Slots nach dem Tausch? Wählen Sie zum Beantworten die entsprechenden Optionen im Antwortbereich aus.
HINWEIS: Jede Korrektur ist einen Punkt wert.

Answer:
Explanation:

Erläuterung:
Das Austauschen der Slots bedeutet, dass die Ziel-Slot-Website-URL den Quell-Slot-Code mit den Ziel-Slot-Einstellungen ausführt.

NEW QUESTION: 3
注：この質問は、同じまたは類似の回答の選択肢を使用する一連の質問の一部です。 回答の選択肢は、シリーズの複数の質問に対して正しいかもしれません。 各質問は、このシリーズの他の質問とは独立しています。 質問に記載されている情報や詳細がその質問に適用されます。
あなたには、銀行システム用のデータベースがあります。 データベースには、預金口座とローン口座をそれぞれ格納する2つのテーブルtblDepositAcctとtblLoanAcctがあります。 どちらの表にも次の列があります。

預金口座とローン口座の合計数を決定する必要があります。
どのTransact-SQL文を実行する必要がありますか？
A. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo L.CustNo
D. SELECT COUNT (DISTINCT COALESCE(D.CustNo, L.CustNo))FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL OR L.CustNo IS NULL
E. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
H. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo=L.CustNoWHERE D.CustNo IS NULL
Answer: F
Explanation:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.

NEW QUESTION: 4
Suppose you were told that scores on an examination were converted to standard scores with a mean = 500, range of 800, and a standard deviation of 100. A person with a score of 600 has performed better than what percent of the persons taking the test?
A. 57 percent
B. 84 percent
C. 97.5 percent
Answer: B
Explanation:
Z = (X - MU)/Standard Deviation Z = (600 - 500)/100 = 1 Area between Z and Mean = .3413
Area to Left of Z = .5 + .3413 = .8413 Percentile Rank = 100*.8413 = 84.13 Therefore, the person has performed better than 84 percent of the people.