Questão de Tecnologia da Informação

1 IP 10.0.1.18.1047 > 10.0.1.200.110: S 59712:59712(0) win 65535 <mss 1460,nop,nop,sackOK> 2 IP 10.0.1.200.110 > 10.0.1.18.1047: S 16141:16141(0) ack 59713 win 65535 <mss 1460,nop,nop,sackOK> 3 IP 10.0.1.18.1047 > 10.0.1.200.110: . ack 1 win 65535 4 IP 10.0.1.200.110 > 10.0.1.18.1047: P 1:19(18) ack 1 win 65535 5 IP 10.0.1.18.1047 > 10.0.1.200.110: . ack 19 win 65517 6 IP 10.0.1.18.1047 > 10.0.1.200.110: P 1:14(13) ack 19 win 65517 7 IP 10.0.1.200.110 > 10.0.1.18.1047: P 19:43(24) ack 14 win 65535 8 IP 10.0.1.18.1047 > 10.0.1.200.110: . ack 43 win 65493 9 IP 10.0.1.18.1047 > 10.0.1.200.110: P 14:27(13) ack 43 win 65493 10 IP 10.0.1.200.110 > 10.0.1.18.1047: . ack 27 win 65535 11 IP 10.0.1.200.110 > 10.0.1.18.1047: P 43:63(20) ack 27 win 65535 12 IP 10.0.1.18.1047 > 10.0.1.200.110: . ack 63 win 65473 13 IP 10.0.1.18.1047 > 10.0.1.200.110: P 27:33(6) ack 63 win 65473 14 IP 10.0.1.200.110 > 10.0.1.18.1047: P 63:91(28) ack 33 win 65535 15 IP 10.0.1.18.1047 > 10.0.1.200.110: F 33:33(0) ack 91 win 65445 16 IP 10.0.1.200.110 > 10.0.1.18.1047: F 91:91(0) ack 33 win 65535 17 IP 10.0.1.18.1047 > 10.0.1.200.110: . ack 92 win 65445 18 IP 10.0.1.200.110 > 10.0.1.18.1047: F 91:91(0) ack 34 win 65535 Considerando o trecho de captura de tráfego acima, julgue os itens subsequentes. O primeiro pacote tem origem no cliente.