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NEW QUESTION: 1
A. #dataTable tbody tr.selected:not(tr:last-child)
B. #dataTable tr.selected:not(tr:last-child)
C. #dataTable > tr.selected:not(tr:last-child)
D. tr:not(tr:last-child).selected < #dataTable
Answer: A
Explanation:
* [attribute]
[target]
Selects all elements with a target attribute
* :not(selector)
not(p)
Selects every element that is not a <p> element
Reference: CSS Selector Reference
NEW QUESTION: 2
When creating a step for a Candidate Selection Workflow, what must you select to map the step to Reports
and Emails? (Choose the best answer.)
A. Reference Model
B. Actions
C. Organization-Location-Job Field elements
D. Statuses
E. Requisition Type
Answer: C
Explanation:
Explanation/Reference:
Explanation:
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 (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE 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 tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
E. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
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:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.
NEW QUESTION: 4
A company plans to implement Power Apps.
For each of the following statements, select Yes if the statement is true. Otherwise, select No.
NOTE: Each correct selection is worth one point.
Answer:
Explanation:
Reference:
https://docs.microsoft.com/en-us/powerapps/maker/