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Exam Code: Introduction-to-Cryptography
Exam Name: WGU Introduction to Cryptography HNO1
Version: V22.75
Q & A: 580 Questions and Answers
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NEW QUESTION: 1
A. 802.1X
B. VLAN
C. MAB
D. interface
E. IP subnet
Answer: A,C
Explanation:
Reference:
https://www.cisco.com/c/dam/en/us/solutions/collateral/borderless-networks/trustsec/C07-730151-00_overview_ page 11
NEW QUESTION: 2
You are creating a Windows Communication Foundation service by using Microsoft .NET Framework 3.5. The service will be hosted in a Console application.
You need to configure the service by using a configuration file other than the default app.config file. Which code segment should you use?
A. Option C
B. Option B
C. Option A
D. Option D
Answer: D
NEW QUESTION: 3
Which statement is true about the number of data sources that can be specified in the Name Service Switch file for each of the data types? (Choose two.)
A. The maximum number that can be specified depends on the system type.
B. A minimum of one can be specified.
C. A maximum of four can be specified.
D. The maximum number that can be specified depends on the data type.
E. A minimum of two can be specified.
Answer: B,D
NEW QUESTION: 4
If Y is a positive integer, does Y have four distinct positive factors?
(1) Y = 8.
(2) Y is a multiplication of two different odd numbers.
A. Either statement BY ITSELF is sufficient to answer the question.
B. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.
D. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
E. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Answer: B
Explanation:
Explanation/Reference:
Explanation:
Statement (1) is sufficient since 8 has the following factors: 1, 2, 4 and 8.
Statement (2) is not sufficient. For example, take 1 and 3, the product is 3, which has only two factors. But if you take 3 and 5, the product is 15 and we have 1,3,5 and 15 as factors of y, and we have four factors.
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