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Reliable C_THR87_2011 Test Sims, Pdf C_THR87_2011 Files | C_THR87_2011 Exam Bootcamp - FreeTorrent

C_THR87_2011

Exam Code: C_THR87_2011

Exam Name: SAP Certified Application Associate - SAP SuccessFactors Variable Pay 2H/2020

Version: V22.75

Q & A: 580 Questions and Answers

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NEW QUESTION: 1
Chris is a senior network administrator. Chris wants to measure the Key Risk Indicator (KRI) to assess the organization. Why is Chris calculating the KRI for his organization? It helps Chris to:
A. Notifies when risk has reached threshold levels
B. Facilitates backward
C. Facilitates post Incident management
D. Identifies adverse events
Answer: A,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 an Azure Cosmos DB account named Account1 that uses the SQL APL Account1 contains a collection named Coll1.
You create a document named doc1 and set the TTL value for the document to 86,400. Thirty-six hours later, you discover that doc1 still exists in Coll1.
You need to ensure that documents are removed from Coll1 when the TTL for the documents expires.
Solution: You set the default TTL value for Coll1 to 129,600.
Does this meet the goal?
A. No
B. Yes
Answer: B

NEW QUESTION: 3
View the Exhibit and examine the structure of the SALES table.
The following query is written to retrieve all those product IDs from the SALES table that have more than
55000 sold and have been ordered more than 10 times.

Which statement is true regarding this SQL statement?

A. It produces an error because COUNT(*) should be specified in the SELECT clause also.
B. It executes successfully but produces no result because COUNT(prod_id) should be used instead of COUNT(*).
C. It produces an error because COUNT(*) should be only in the HAVING clause and not in the WHERE clause.
D. It executes successfully and generates the required result.
Answer: C
Explanation:
Explanation/Reference:
Explanation:
Restricting Group Results with the HAVING Clause
You use the HAVING clause to specify the groups that are to be displayed, thus further restricting the groups on the basis of aggregate information.
In the syntax, group_condition restricts the groups of rows returned to those groups for which the specified condition is true.
The Oracle server performs the following steps when you use the HAVING clause:
1. Rows are grouped.
2. The group function is applied to the group.
3. The groups that match the criteria in the HAVING clause are displayed.
The HAVING clause can precede the GROUP BY clause, but it is recommended that you place the GROUP BY clause first because it is more logical. Groups are formed and group functions are calculated before the HAVING clause is applied to the groups in the SELECT list.
Note: The WHERE clause restricts rows, whereas the HAVING clause restricts groups.

NEW QUESTION: 4
What is the Limitations of ALV for SAP HANA? There are 3 correct answers to this question.
A. All unit/ currency restricted to 10,000,00 records
B. Restricted to 10,000 records
C. Without currency/unit consideration
D. Restricted to 5000 records / unit
E. No unit/ currency split if aggregating amounts/ currency values
Answer: B,C,E

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