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NEW QUESTION: 1
A company uses SharePoint Online and SharePoint Server 2016 on-premises.
You need to configure two-way federated search between SharePoint Online and SharePoint Server on-premises.
How should you configure federated search? To answer, drag the appropriate option to the correct result source and remote providers. Each option may be used once, more than once, or not at all. You may need to drag the split bar between panes or scroll to view content.
Note: Each correct selection is worth one point.

Answer:
Explanation:

Explanation

NEW QUESTION: 2
Some additional server-side JavaScript @Functions have been provided via the Extension Library for presenting messages to users. Which three @Formulas can be used? (Choose three.)
A. @ErrorMessage
B. @FailureMessage
C. @WarningMessage
D. @Prompt
E. @InfoMessage
F. @IsError
Answer: A,C,E

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 tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
D. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
E. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL
F. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALL SELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
H. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
Answer: F
Explanation:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.