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NEW QUESTION: 1
A. Option A
B. Option E
C. Option C
D. Option B
E. Option D
Answer: A,B
NEW QUESTION: 2
To work on pat_r2 branch, what is line 3 in view config specs? 1. element * CHECKEDOUT
2. element * /main/r2_int/pat_r2/LATEST 3. 4. element * R1 kbranch r2_int4. element * R1 ?kbranch r2_int 5. element * /main/LATEST
A. element * /main/r2_int kbranch pat_r2element * /main/r2_int ?kbranch pat_r2
B. element * /main/LATEST kbranch pat_r2element * /main/LATEST ?kbranch pat_r2
C. element * /main/r2_int/LATEST
D. element * /main/r2_int/LATEST kbranch pat_r2element * /main/r2_int/LATEST ?kbranch pat_r2
Answer: D
NEW QUESTION: 3
An administrator needs to configure virtual printing to allow users to access locally attached printers.
What must the administrator do to enable this functionality?
A. Import the vdm_client.adm template to the Active Directory server and enable local printing.
B. Ensure virtual printing is enabled in the View Agent installation options.
C. Install printer drivers in the base image for the pool.
D. Register the TPVMGPoACmap.dll file on the base image for the pool.
Answer: B
NEW QUESTION: 4
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 tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
B. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
C. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
D. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
E. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo L.CustNo
G. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo=L.CustNoWHERE D.CustNo IS NULL
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
Answer: D
Explanation:
Would list the customers with duplicates, which would equal the number of accounts.