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Exam NS0-ASM-ASE3 Consultant, Network Appliance Exam NS0-ASM-ASE3 Tutorials | New NS0-ASM-ASE3 Test Practice - Sugakumaster
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Exam Code: NS0-ASM-ASE3
Exam Name: Accredited Service Engineer Level 3 ExamCertification
Version: V16.75
Q & A: 400 Questions and Answers
NS0-ASM-ASE3 Free Demo download
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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. Amazon Cognito identity pools
B. Amazon Cognito user pools
C. AWS Directory Service
D. AWS IAM
Answer: A
NEW QUESTION: 2
Which software-defined storage (SDS) product is optimized for virtual deployments?
A. HPE MSA
B. HPE StoreServ
C. HPE StoreVirtual
D. HPE XP
Answer: C
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(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
B. 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
C. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
D. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
E. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT
CustNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL
H. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
Answer: E
Explanation:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.
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