Given Packet Capture Following Scan Following Likely Inferred Scan S Output 1921681115 Ho Q43812340
Given a packet capture of the following scan:
![nmap -sX 192.168.1.55 - 22,80, 445 45 33.105540 192.168.1.115 192.168.1.55 TCP 54 39007 -> 80 (FIN, PSH, URG] Seq=1 Win=1024](https://media.cheggcdn.com/media/665/6654986b-a6f9-4dc1-bc8d-708e609f1fdd/phpm2WTWY.png)
Which of the following should MOST likely be inferred on thescan’s output?
A. 192.168.1.115 is hosting a web server.
B. 192.168.1.55 is hosting a web server.
C. 192.168.1.55 is a Linux server.
D. 192.168.1.55 is a file server.
nmap -sX 192.168.1.55 – 22,80, 445 45 33.105540 192.168.1.115 192.168.1.55 TCP 54 39007 -> 80 (FIN, PSH, URG] Seq=1 Win=1024 Urg=0 Len=0 46 33.106599 192.168.1.115 192.168.1.55 TCP 54 39007 -> 445 [FIN, PSH, URG] Seq=1 Win=1024 Urg=0 Len=0 47 33.107672 192.168.1.115 192.168.1.55 TCP 54 39007 -> 22 (FIN, PSH, URG] Seq=1 Win=1024 Urg=0 Len=0 48 33.108730 192.168.1.55 192.168.1.115 TCP 54 445 -> 39007 [RST, ACK) Seq=1 Ack=2 Urg=0 Len=0 49 33.108972 192.168.1.55 192.168.1.115 TCP 54 22 -> 39007 (RST, ACK] Seq=1 Ack=2 Urg=0 Len=0 50 34.207377 192.168.1.115 192.168.1.55 TCP 54 39008 -> 80 (FIN, PSH, URG) Seq=1 Win=1024 Urg=0 Len=0 Show transcribed image text nmap -sX 192.168.1.55 – 22,80, 445 45 33.105540 192.168.1.115 192.168.1.55 TCP 54 39007 -> 80 (FIN, PSH, URG] Seq=1 Win=1024 Urg=0 Len=0 46 33.106599 192.168.1.115 192.168.1.55 TCP 54 39007 -> 445 [FIN, PSH, URG] Seq=1 Win=1024 Urg=0 Len=0 47 33.107672 192.168.1.115 192.168.1.55 TCP 54 39007 -> 22 (FIN, PSH, URG] Seq=1 Win=1024 Urg=0 Len=0 48 33.108730 192.168.1.55 192.168.1.115 TCP 54 445 -> 39007 [RST, ACK) Seq=1 Ack=2 Urg=0 Len=0 49 33.108972 192.168.1.55 192.168.1.115 TCP 54 22 -> 39007 (RST, ACK] Seq=1 Ack=2 Urg=0 Len=0 50 34.207377 192.168.1.115 192.168.1.55 TCP 54 39008 -> 80 (FIN, PSH, URG) Seq=1 Win=1024 Urg=0 Len=0
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Answer to Given a packet capture of the following scan: Which of the following should MOST likely be inferred on the scan’s outp…
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