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Pegasystems PEGACPBA24V1 - Certified Pega Business Architect 24 Exam Braindumps

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  • Exam Code:PEGACPBA24V1
  • Exam Name:Certified Pega Business Architect 24 Exam Exam
  • Total Questions:276 Questions and Answers
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NEW QUESTION: 1
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 customers who have either deposit accounts or loan accounts, but not both types of 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 D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
C. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
D. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM 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 (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
G. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
Answer: F
Explanation:
Explanation
SQL Server provides the full outer join operator, FULL OUTER JOIN, which includes all rows from both tables, regardless of whether or not the other table has a matching value.
Consider a join of the Product table and the SalesOrderDetail table on their ProductID columns. The results show only the Products that have sales orders on them. The ISO FULL OUTER JOIN operator indicates that all rows from both tables are to be included in the results, regardless of whether there is matching data in the tables.
You can include a WHERE clause with a full outer join to return only the rows where there is no matching data between the tables. The following query returns only those products that have no matching sales orders, as well as those sales orders that are not matched to a product.
USE AdventureWorks2008R2;
GO
-- The OUTER keyword following the FULL keyword is optional.
SELECT p.Name, sod.SalesOrderID
FROM Production.Product p
FULL OUTER JOIN Sales.SalesOrderDetail sod
ON p.ProductID = sod.ProductID
WHERE p.ProductID IS NULL
OR sod.ProductID IS NULL
ORDER BY p.Name ;
References: https://technet.microsoft.com/en-us/library/ms187518(v=sql.105).aspx

NEW QUESTION: 2
Siehe Ausstellung.

Der New Yorker Router ist mit statischen Routen konfiguriert, die auf die Standorte Atlanta und Washington verweisen. Welche beiden Aufgaben müssen ausgeführt werden, damit sich die Serial0 / 0/0-Schnittstellen der Router in Atlanta und Washington gegenseitig erreichen können?
(Wähle zwei.)
A. Konfigurieren Sie den Befehl ipv6 route 2012 :: / 126 s0 / 0/0 auf dem Atlanta-Router
B. Konfigurieren Sie den Befehl ipv6 route 2023 :: / 126 2012 :: 1 auf dem Atlanta-Router
C. Konfigurieren Sie den Befehl ipv6 route 2012 :: / 126 2023 :: 2 auf dem Washington-Router
D. Konfigurieren Sie den Befehl ipv6 route 2012 :: / 126 2023 :: 1 auf dem Washington-Router
E. Konfigurieren Sie den Befehl ipv6 route 2023 :: / 126 2012 :: 2 auf dem Atlanta-Router
Answer: C,E

NEW QUESTION: 3
Drag and Drop Question
Drag and drop the Cisco ISE portal types from the left onto the correct functionalities on the right.

Answer:
Explanation:


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