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SAP Certified Associate - SAP S/4HANA Cloud Private Edition, Asset Management

  • Exam Number/Code : C-S43-2023
  • Exam Name : SAP Certified Associate - SAP S/4HANA Cloud Private Edition, Asset Management
  • Questions and Answers : 213 Q&As
  • Update Time: 2019-01-10
  • Price: $ 99.00 $ 39.00

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NEW QUESTION: 1
Refer to the exhibit.

You have a Contrail providing cloud services to Customer A.
Customer A wants to add a packet capture service to their network to analyze traffic flowing to and from the Internet.
Referring to the exhibit, which statement is true?
A. The packet sniffing service must be performed on the edge routers facing the Internet.
B. The packet sniffing service must be enabled on the vRouter directly attached to the customer's VM.
C. The packet sniffing service can be implemented on any physical host in the cloud environment.
D. The packet sniffing service can be implemented on any switch within the underlay fabric.
Answer: B

NEW QUESTION: 2
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 run a query to find the total number of customers who have both deposit and loan accounts.
Which Transact-SQL statement should you run?
A. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
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 tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
D. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
E. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
F. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
G. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
Answer: F
Explanation:
Explanation
The SQL INTERSECT operator is used to return the results of 2 or more SELECT statements. However, it only returns the rows selected by all queries or data sets. If a record exists in one query and not in the other, it will be omitted from the INTERSECT results.
References: https://www.techonthenet.com/sql/intersect.php

NEW QUESTION: 3
A customer requires different retention categories for items in users' mailbox folders. Which two options could an administrator use to assign different retention categories to items in the specified folders? (Select two.)
A. Exchange Managed folders
B. Enterprise Vault Archive settings
C. Enterprise Vault Desktop policy
D. Exchange Mailbox policy
E. Enterprise Vault Policy Manager
Answer: A,E