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Exam 312-49 All Questions

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Exam 312-49 topic 1 question 61 discussion

Actual exam question from ECCouncil's 312-49
Question #: 61
Topic #: 1
[All 312-49 Questions]

What method of computer forensics will allow you to trace all ever-established user accounts on a Windows 2000 sever the course of its lifetime?

  • A. forensic duplication of hard drive
  • B. analysis of volatile data
  • C. comparison of MD5 checksums
  • D. review of SIDs in the Registry
Show Suggested Answer Hide Answer
Suggested Answer: D 🗳️

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Pet
Highly Voted 5 years, 5 months ago
Answer shd be D. review of SIDs in the Registry
upvoted 9 times
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haymaths
Highly Voted 5 years, 3 months ago
Answer is D. Option C is totally out of it
upvoted 7 times
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Bennoli13
Most Recent 5 months, 2 weeks ago
To trace all user accounts that have ever been established on a Windows 2000 server over its lifetime, you would need to review the Security Identifiers (SIDs) in the Registry. Each user account on a Windows system is assigned a unique SID, which is stored in the Registry. By examining these SIDs, you can identify all user accounts that have been created on the system. Therefore, the correct answer is: D. review of SIDs in the Registry
upvoted 1 times
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tej0n
1 year, 2 months ago
Selected Answer: D
SID registry identifies all history user account in Windows.
upvoted 1 times
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simbai
1 year, 5 months ago
The Security Identifier (SID) is a unique identifier assigned to each user account in Windows. By reviewing the SIDs in the Registry, a forensic investigator can trace all ever-established user accounts on a Windows 2000 server over the course of its lifetime
upvoted 1 times
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jordy55
2 years ago
Selected Answer: D
Defintly D
upvoted 1 times
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fhramke
2 years, 5 months ago
Selected Answer: D
Answer shd be D. review of SIDs in the Registry
upvoted 1 times
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ZeroDay
4 years, 6 months ago
The answere is definitely The review of SIDs in the Reg. D
upvoted 4 times
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ireen
5 years, 3 months ago
Answer should be D. As user account are assigned a unique SID and SID are not reused.
upvoted 5 times
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Community vote distribution
A (35%)
C (25%)
B (20%)
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