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Exam CPTIA Consultant, CREST Exam CPTIA Tutorials | New CPTIA Test Practice - Sugakumaster
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Exam Code: CPTIA
Exam Name: CREST Practitioner Threat Intelligence AnalystCertification
Version: V16.75
Q & A: 400 Questions and Answers
CPTIA Free Demo download
About CREST CPTIA Exam
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NEW QUESTION: 1
開発者は、新しいユーザーが新しいユーザーアカウントを登録および作成できるようにするアプリケーションを構築したいと考えています。また、アプリケーションは、ソーシャルメディアアカウントを持つユーザーがソーシャルメディアの資格情報を使用してログインできるようにする必要があります。
これらの要件を満たすために使用できるAWSサービスまたは機能はどれですか?
A. AWS IAM
B. Amazon Cognito user pools
C. AWS Directory Service
D. Amazon Cognito identity pools
Answer: D
NEW QUESTION: 2
Which software-defined storage (SDS) product is optimized for virtual deployments?
A. HPE StoreVirtual
B. HPE XP
C. HPE MSA
D. HPE StoreServ
Answer: A
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 D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
B. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
C. 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
D. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
E. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT
CustNoFROM tblLoanAcct) R
G. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT 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: F
Explanation:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.
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