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Exam Code: 2V0-72.22
Exam Name: Professional Develop VMware Spring
Vendor: VMware

60 Questions & Answers
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NEW QUESTION: 1
A customer has a mainframe running z/OS and Linux and wishes to use both an IBM DS8886 as well as IBM Spectrum Virtualize.
Which combination fulfills the customer requirements?
A. CKD and 8-way multipathing
B. FB and ALUA
C. 8-way multipathing and ALUA
D. FB and CKD
Answer: D

NEW QUESTION: 2
Note: This question is part of a series of questions that present the same scenario. Each question in the series contains a unique solution that might meet the stated goals. Some question sets might have more than one correct solution, while others might not have a correct solution.
After you answer a question in this section, you will NOT be able to return to it. As a result, these questions will not appear in the review screen.
You have a Microsoft SQL server that has Data Quality Services (DQS) installed.
You need to review the completeness and the uniqueness of the data stored in the matching policy.
Solution: You modify the weight of the domain in the matching rule.
Does this meet the goal?
A. Yes
B. No
Answer: A
Explanation:
Explanation/Reference:
Explanation:
Use a matching rule, and use completeness and uniqueness data to determine what weight to give a field in the matching process.
If there is a high level of uniqueness in a field, using the field in a matching policy can decrease the matching results, so you may want to set the weight for that field to a relatively small value. If you have a low level of uniqueness for a column, but low completeness, you may not want to include a domain for that column.
References: https://docs.microsoft.com/en-us/sql/data-quality-services/create-a-matching-policy?view=sql- server-2017

NEW QUESTION: 3
The following SAS program is submitted:
data combine; prefix='505'; middle='6465 '; end='09090'; <insert statement here>; run;
Which statement successfully completes the program so that TOTAL has a value of 505-646509090?
A. total = prefix!!'-'!! left(middle)!!'-'!! end;
B. total = catx('-', prefix, middle, end);
C. total = cat('-', prefix, middle, end);
D. total = prefix !!'-'!! middle ''!!'-'!! end;
Answer: B


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