Exam Code: 2V0-71.23
Exam Name: VMware Tanzu for Kubernetes Operations Professional
Version: V13.25
Q & A: 72 Questions and Answers
2V0-71.23 Free Demo download
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NEW QUESTION: 1
You are a consultant to Mary's company for her Google AdWords account. Mary would like to
target ads for YouTube, but she does not want to create a new YouTube account for advertising. Since you are the Google AdWords and YouTube advertising consultant, what advice would you offer to Mary in this scenario?
A. Mary does not need to open a YouTube account as Google AdWords already appear on YouT ube.
B. Mary can use your YouTube account for her company in order to get her ads on YouTube.
C. Mary will need to create a YouTube account for her company in order to purchase ads on Y ouTube.
D. Mary does not need to open a YouTube account; she can use her Google AdWords account to place certain ad types directly on YouTube.
Answer: D
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 run a query to find the total number of customers who have both deposit and loan accounts.
Which Transact-SQL statement should you run?
A. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE
C. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT
CustNoFROM tblLoanAcct) R
D. CustNo = L.CustNo
E. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo
F. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT
CustNoFROM tblLoanAcct) R
G. CustNo IS NULL
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 AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM 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
K. CustNo
Answer: I
Explanation:
The SQL INTERSECT operator is used to return the results of 2 or more SELECT statements. However, it only returns the rows selected by all queries or data sets. If a record exists in one query and not in the other, it will be omitted from the INTERSECT results.
References: https://www.techonthenet.com/sql/intersect.php
NEW QUESTION: 3
Given a packet capture of the following scan:
Which of the following should MOST likely be inferred on the scan's output?
A. 192.168.1.55 is hosting a web server.
B. 192.168.1.55 is a Linux server.
C. 192.168.1.55 is a file server.
D. 192.168.1.115 is hosting a web server.
Answer: C
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