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Exam Code: 1Z0-996-21

Exam Name: Oracle Utilities Customer Cloud Service 2021 Implementation Essentials

Certification Provider: Oracle

Related Certification: Oracle Utilities Customer Cloud Service 2021 Implementation Essentials

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NEW QUESTION: 1
What is the differences between TCAM and the MAC address table?
A. The MAC address table is contained in CAM ACL and QoS information is stored in TCAM
B. Router prefix lookups happens in CAM. MAC address table lookups happen in TCAM.
C. The MAC address table supports partial matches. TCAM requires an exact match
D. TCAM is used to make Layer 2 forwarding decisions CAM is used to build routing tables
Answer: A
Explanation:
Explanation
https://community.cisco.com/t5/networking-documents/cam-content-addressable-memory-vs-tcam-ternary-cont

NEW QUESTION: 2
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 AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT
CustNoFROM tblLoanAcct) R
C. CustNo IS NULL
D. CustNo
E. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE
F. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo
G. SELECT COUNT (DISTINCT COALESCE(D.CustNo, L.CustNo))FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL OR
H. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN
tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL
I. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT
CustNoFROM tblLoanAcct) R
J. CustNo = L.CustNo
K. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT
CustNoFROM tblLoanAcct) R
Answer: I
Explanation:
Would list the customers with duplicates, which would equal the number of accounts.


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