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IBM S2000-025 - IBM AIX v7.3 Administrator Specialty Exam Braindumps

IBM S2000-025 - IBM AIX v7.3 Administrator Specialty Exam Braindumps

  • Certification Provider:IBM
  • Exam Code:S2000-025
  • Exam Name:IBM AIX v7.3 Administrator Specialty Exam Exam
  • Total Questions:276 Questions and Answers
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NEW QUESTION: 1
A company hires outside security experts to evaluate the security status of the corporate network. All of the company's IT resources are outdated and prone to crashing. The company requests that all testing be performed in a way which minimizes the risk of system failures. Which of the following types of testing does the company want performed?
A. White box testing
B. Vulnerability scanning
C. WAF testing
D. Penetration testing
Answer: B
Explanation:
Explanation/Reference:
Explanation:
Vulnerability scanning has minimal impact on network resource due to the passive nature of the scanning.
A vulnerability scan is the process of scanning the network and/or I.T. infrastructure for threats and vulnerabilities. The threats and vulnerabilities are then evaluated in a risk assessment and the necessary actions taken to resolve and vulnerabilities. A vulnerability scan scans for known weaknesses such as missing patches or security updates.
A vulnerability scan is the automated process of proactively identifying security vulnerabilities of computing systems in a network in order to determine if and where a system can be exploited and/or threatened.
While public servers are important for communication and data transfer over the Internet, they open the door to potential security breaches by threat agents, such as malicious hackers.
Vulnerability scanning employs software that seeks out security flaws based on a database of known flaws, testing systems for the occurrence of these flaws and generating a report of the findings that an individual or an enterprise can use to tighten the network's security.

NEW QUESTION: 2
Which of the following is an asymmetric encryption algorithm?
A. Diffie-Hellman
B. Blowfish
C. RC5
D. RC4
Answer: A

NEW QUESTION: 3
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.0.255.255.255.254
B. Permit 10.0.0.1.0.0.0.254
C. Permit 10.0.0.1.0.0.0.0
D. Permit 10.0.0.0.0.0.0.1
Answer: B
Explanation:
Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's are I CARE. So now let's
analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of
the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care
about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16;
maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we
only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd
numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even
numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with
a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered
hosts in the 10.0.0.0/24 subnet.

NEW QUESTION: 4
コントローラの「クライアントの監視」セクションでクライアントを見ると、次の情報のどれが表示されていませんか?
A. 役割
B. クライアント無線の出力電力
C. 年齢
D. MACアドレス
E. 認証方法
Answer: B

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