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NEW QUESTION: 1
Cisco DNA Centerでデバイスをプロビジョニングすると、エンジニアは「デバイスを選択できません。テンプレートと互換性がありません」というエラーメッセージを表示します。
エラーの理由は何ですか？
A. テンプレートの構成が正しくありません。
B. テンプレートへの変更はコミットされませんでした。
C. テンプレートのソフトウェアバージョンがデバイスのソフトウェアバージョンと異なります。
D. テンプレートのフィルタリングに使用されたタグがデバイスタグと一致しません。
Answer: D
Explanation:
Explanation
If you use tags to filter the templates, you must apply the same tags to the device to which you want to apply the templates. Otherwise, you get the following error during provisioning: Cannot select the device. Not compatible with template.
NEW QUESTION: 2
Ein Unternehmen hostet eine Standardinstallation von Microsoft SQL Server 2017 auf einem Linux-basierten Server.
Sie stellen fest, dass sich die Leistung des Servers verschlechtert hat. Prozesse zum Zuweisen von Tabellen weisen eine hohe Latenz auf.
Sie müssen eine Lösung empfehlen, um die Leistung beim Zuweisen von Tabellen zu verbessern.
Was solltest du empfehlen?
A. Installieren Sie SQL Server unter Linux neu.
B. Konfigurieren Sie zusätzliche Tempdb-Datendateien.
C. Legen Sie ein Speicherlimit für den SQL Server fest.
D. Erhöhen Sie das Limit für den virtuellen Adressraum.
Answer: C
Explanation:
Explanation
References:
https://www.itprotoday.com/microsoft-sql-server/top-10-sql-server-performance-tuning-tips
NEW QUESTION: 3
Note: This question is part of a series of questions that use the same or similar answer choices. An answer choice may be correct for more than one question in the series. Each question is independent of the other questions in this series. Information and details provided in a question apply to that question.
You have a database for a banking system. The database has two tables named tblDepositAcct and tblLoanAcct that store deposit and loan accounts, respectively. Both tables contain the following columns:
You need to determine the total number of deposit and loan accounts.
Which Transact-SQL statement should you run?
A. 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
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
D. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
E. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
F. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL
Answer: G
Explanation:
Would list the customers with duplicates, which would equal the number of accounts.