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Oracle 1Z0-1050-23 - Oracle Payroll Cloud 2023 Implementation Professional Exam Braindumps

Oracle 1Z0-1050-23 - Oracle Payroll Cloud 2023 Implementation Professional Exam Braindumps

  • Certification Provider:Oracle
  • Exam Code:1Z0-1050-23
  • Exam Name:Oracle Payroll Cloud 2023 Implementation Professional Exam Exam
  • Total Questions:276 Questions and Answers
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NEW QUESTION: 1
How does HP Requirements Management differentiate itself from other solutions? (Select three.)
A. It offers performance modeling and scalability.
B. It provides both requirements definition and management capabilities.
C. It can import requirements defined in other tools, such as Word and BluePrint.
D. Itidentifies and diagnoses bottlenecks through advanced analysis, monitoring, and diagnostics.
E. It is fully integrated with HP ALM to ensure that requirements are visible to all the stakeholders in the application lifecycle.
F. It offers deployment throughout the applications lifecycle.
Answer: B,D,E

NEW QUESTION: 2
Given:
package com.sun.scjp;
public class Geodetics {
public static final double DIAMETER = 12756.32; // kilometers
}
Which two correctly access the DIAMETER member of the Geodetics class? (Choose two.)
A. package com.sun.scjp;
public class TerraCarta {
public double halfway() { return DIAMETER/2.0; }
}
B. import static com.sun.scjp.Geodetics;
public class TerraCarta{
public double halfway() { return DIAMETER/2.0; }
}
C. import com.sun.scjp.Geodetics;
public class TerraCarta {
public double halfway() { return Geodetics.DIAMETER/2.0; }
}
D. import static com.sun.scjp.Geodetics.*;
public class TerraCarta {
public double halfway() { return DIAMETER/2.0; }
}
Answer: C,D

NEW QUESTION: 3
You plan to design an application that temporarily stores data in a SQL Azure database. You need to identify which types of database objects can be used to store data for the application. The solution must ensure that the application can make changes to the schema of a temporary object during a session. Which type of objects should you identify?
A. Temporary tables
B. Temporary stored procedures
C. Common table expressions (CTEs)
D. Table variables
Answer: A
Explanation:
According to these references, the answer looks correct.
References: http://msdn.microsoft.com/en-us/library/ms175972.aspx http://msdn.microsoft.com/en-us/library/ms189084.aspx http://msdn.microsoft.com/en-us/library/ms175010.aspx http://msdn.microsoft.com/en-us/library/bb510489.aspx http://msdn.microsoft.com/en-us/library/ms187926.aspx http://zacksfiasco.com/post/2010/01/21/SQL-Server-Temporary-Stored-Procedures.aspx

NEW QUESTION: 4
Which type of algorithm is considered to have the highest strength per bit of key length of any of the asymmetric algorithms?
A. Rivest, Shamir, Adleman (RSA)
B. Advanced Encryption Standard (AES)
C. El Gamal
D. Elliptic Curve Cryptography (ECC)
Answer: D
Explanation:
The Answer: "Elliptic Curve Cryptography (ECC)". This type of cryptography is based on the complex mathematics of elliptic curves. These algorithms are advantageous for their speed and strength.
The other answers are not correct because:
"Rivest, Shamir, Adleman (RSA)" is incorrect because RSA is a "traditional" asymmetric algorithm. While it is reasonably strong, it is not considered to be as strong as ECC based systems.
"El Gamal" is incorrect because it is also a "traditional" asymmetric algorithm and not considered as strong as ECC based systems.
"Advanced Encryption Standard (AES)" is incorrect because the question asks specifically about asymmetric algorithms and AES is a symmetric algorithm.
References:
Official ISC2 Guide page: 258
All in One Third Edition page: 638
The RSA Crypto FAQ: http://www.rsa.com/rsalabs/node.asp?id=2241

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