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NEW QUESTION: 1
Refer to the exhibit.

An engineer is configuring the New Your router to reach the Lo1 interface of the Atlanta router using interface Se0/0/0 as the primary path.
Which two commands must be configured on the New York router so that it can reach the Lo1 interface of the Atlanta router via Washington when the link between New York and Atlanta goes down? (Choose two.)
A. Ipv6 route 2000::1/128 2023::2 5
B. Ipv6 route 2000::1/128 2012::1 5
C. Ipv6 route 2000::1/128 2023::3 5
D. Ipv6 route 2000::1/128 2012::2
E. Ipv6 route 2000::1/128 2012::1
Answer: C,E

NEW QUESTION: 2
Scenario: A Citrix Administrator needs to integrate LDAP for Citrix ADC system administration using current Active Directory (AD) groups. The administrator created the group on the Citrix ADC, exactly matching the group name in LDAP.
What can the administrator bind to specify the permission level and complete the LDAP configuration?
A. Users to the group on the Citrix ADC
B. A command policy to the group
C. A nested group to the new group
D. An authentication, authorization, and auditing (AAA) action to the group
Answer: B

NEW QUESTION: 3



A. CustNo
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo
D. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
E. CustNoWHERE D.CustNo IS NULL
F. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo
G. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
I. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
J. 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
Answer: J
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