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Exam Code: TCA-Tibco-BusinessWorks
Exam Name: TIBCO BusinessWorks Certified Associates Exam
Version: V22.75
Q & A: 580 Questions and Answers
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NEW QUESTION: 1
Which two options describe services that are provided by TCP? (Choose two.)
A. best-effort packet delivery
B. binary translation
C. session multiplexing
D. EtherTypes identify destination services
E. connection-oriented
Answer: C,E
Explanation:
Explanation/Reference:
Explanation:
"TCP is a connection-oriented protocol."
http://www.diffen.com/difference/TCP_vs_UDP
NEW QUESTION: 2
Acceptance tests of user stories are specified by the:
A. Tester after the team has finished the coding.
B. Customer as late as possible during the iteration.
C. Customer at the start of the iteration.
D. Development team at the start of the iteration.
Answer: C
NEW QUESTION: 3
A. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo
D. CustNoWHERE D.CustNo IS NULL
E. CustNo
F. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo
G. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
H. 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
I. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
J. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
Answer: F
Explanation:
The RIGHT JOIN keyword returns all records from the right table (table2), and the matched records from the left table (table1). The result is NULL from the left side, when there is no match.
References: https://www.w3schools.com/sql/sql_join_right.asp
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