Pass4Test vous promet de vous aider à passer le test EC-COUNCIL 312-49, vous pouvez télécharger maintenant les Q&As partielles de test EC-COUNCIL 312-49 en ligne. Il y a encore la mise à jour gratuite pendant un an pour vous. Si vous malheureusement rater le test, votre argent sera 100% rendu.
Le test EC-COUNCIL 312-49 est populaire dans l'Industrie IT. Il y a beaucoup de professionnels IT veulent ce passport de IT. Votre vie et salaire sera améliorée avec ce Certificat. Vous aurez une meilleure assurance.
Pass4Test est un site d'offrir l'outil de formation convenable pour les candidats de test Certification IT. Le produit de Pass4Test peut aider les candidats à économiser les temps et les efforts. L'outil de formation est bien proche que test réel. Vous allez réussir le test 100% avec l'aide de test simulation de Pass4Test. C'est une bonne affaire à prendre le Certificat IT en coûtant un peu d'argent. N'hésitez plus d'ajouter l'outil de formation au panier.
Code d'Examen: 312-49
Nom d'Examen: EC-COUNCIL (Computer Hacking Forensic Investigator )
Questions et réponses: 150 Q&As
Le test EC-COUNCIL 312-49 est une examination de techniques professionnelles dans l'Industrie IT. Pass4Test est un site qui peut vous aider à réussir le test EC-COUNCIL 312-49 rapidement. Si vous utiliser l'outil de formation avant le test, vous apprendrez tous essences de test Certification EC-COUNCIL 312-49.
312-49 Démo gratuit à télécharger: http://www.pass4test.fr/312-49.html
NO.1 The offset in a hexadecimal code is:
A. The last byte after the colon
B. The 0x at the beginning of the code
C. The 0x at the end of the code
D. The first byte after the colon
Answer: B
EC-COUNCIL examen 312-49 certification 312-49
NO.2 A suspect is accused of violating the acceptable use of computing resources, as he has visited
adult websites and downloaded images. The investigator wants to demonstrate that the suspect
did indeed visit these sites. However, the suspect has cleared the search history and emptied the
cookie cache. Moreover, he has removed any images he might have downloaded. What can the
investigator do to prove the violation? Choose the most feasible option.
A. Image the disk and try to recover deleted files
B. Seek the help of co-workers who are eye-witnesses
C. Check the Windows registry for connection data (You may or may not recover)
D. Approach the websites for evidence
Answer: A
certification EC-COUNCIL 312-49 312-49 certification 312-49
NO.3 It takes _____________ mismanaged case/s to ruin your professional reputation as a computer
forensics examiner?
A. by law, three
B. quite a few
C. only one
D. at least two
Answer: C
certification EC-COUNCIL 312-49 examen 312-49
NO.4 What does the superblock in Linux define?
A. file system names
B. available space
C. location of the first inode
D. disk geometry
Answer: B, C, D
EC-COUNCIL examen 312-49 312-49 312-49
NO.5 In a computer forensics investigation, what describes the route that evidence takes from the time
you find it until the case is closed or goes to court?
A. rules of evidence
B. law of probability
C. chain of custody
D. policy of separation
Answer: C
EC-COUNCIL certification 312-49 312-49 examen certification 312-49
NO.6 A honey pot deployed with the IP 172.16.1.108 was compromised by an attacker . Given below is
an excerpt from a Snort binary capture of the attack. Decipher the activity carried out by the
attacker by studying the log. Please note that you are required to infer only what is explicit in the
excerpt. (Note: The student is being tested on concepts learnt during passive OS fingerprinting,
basic TCP/IP connection concepts and the ability to read packet signatures from a sniff dump.)
03/15-20:21:24.107053 211.185.125.124:3500 -> 172.16.1.108:111
TCP TTL:43 TOS:0x0 ID:29726 IpLen:20 DgmLen:52 DF
***A**** Seq: 0x9B6338C5 Ack: 0x5820ADD0 Win: 0x7D78 TcpLen: 32
TCP Options (3) => NOP NOP TS: 23678634 2878772
=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=
03/15-20:21:24.452051 211.185.125.124:789 -> 172.16.1.103:111
UDP TTL:43 TOS:0x0 ID:29733 IpLen:20 DgmLen:84
Len: 64
01 0A 8A 0A 00 00 00 00 00 00 00 02 00 01 86 A0 ................
00 00 00 02 00 00 00 03 00 00 00 00 00 00 00 00 ................
00 00 00 00 00 00 00 00 00 01 86 B8 00 00 00 01 ................
00 00 00 11 00 00 00 00 ........
=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=+=
03/15-20:21:24.730436 211.185.125.124:790 -> 172.16.1.103:32773
UDP TTL:43 TOS:0x0 ID:29781 IpLen:20 DgmLen:1104
Len: 1084
47 F7 9F 63 00 00 00 00 00 00 00 02 00 01 86 B8 G..c............
00 00 00 01 00 00 00 01 00 00 00 01 00 00 00 20 ...............
3A B1 5E E5 00 00 00 09 6C 6F 63 61 6C 68 6F 73 :.
没有评论:
发表评论