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NEW QUESTION: 1
HOTSPOT


Answer:
Explanation:

NEW QUESTION: 2
BYODソリューションでは、どの2つの主要なセキュリティ問題を緩和できますか？ （2つ選択）
A. 信頼できる企業ネットワークへのアクセスを保護する。
B. デバイスのパッチを適用するスケジュール
C. デバイスの遅延と在庫
D. 適用可能なポリシーへの準拠
E. 公衆Wi-Fiネットワークへの接続
Answer: A,D

NEW QUESTION: 3
-- Exhibit - -- Exhibit -

You are designing a network for NSX and your customer has stated that virtual machine tra ffic needs to span the virtual and physical space.
Based on the diagram, where should this requirement be configured?
A. Location A
B. Location C
C. Locations B and C
D. Location B
Answer: D

NEW QUESTION: 4
The LTM device is configured to provide load balancing to a set of web servers that implement access control lists (ACL) based on the source IP address of the client. The ACL is at the network level and the web server is configured to send a TCP reset back to the client if it is NOT permitted to connect.
The virtual server is configured with the default OneConnect profile.
The ACL is defined on the web server as:
Permit: 192.168.136.0/24
Deny: 192.168.116.0/24
The packet capture is taken of two individual client flows to a virtual server with IP address 192.168.136.100.
Client A - Src IP 192.168.136.1 - Virtual Server 192.168.136.100:
Clientside:
09:35:11.073623 IP 192.168.136.1.55684 > 192.168.136.100.80: S 869998901:869998901(0) win 8192 <mss
1460,nop,wscale 2,nop,nop,sackOK>
09:35:11.073931 IP 192.168.136.100.80 > 192.168.136.1.55684: S 2273668949:2273668949(0) ack
869998902 win 4380 <mss 1460,nop,wscale 0,sackOK,eol>
09:35:11.074928 IP 192.168.136.1.55684 > 192.168.136.100.80: . ack 1 win 16425
09:35:11.080936 IP 192.168.136.1.55684 > 192.168.136.100.80: P 1:299(298) ack 1 win 16425
09:35:11.081029 IP 192.168.136.100.80 > 192.168.136.1.55684: . ack 299 win 4678 Serverside:
09:35:11.081022 IP 192.168.136.1.55684 > 192.168.116.128.80: S 685865802:685865802(0) win 4380 <mss
1460,nop,wscale 0,sackOK,eol>
09:35:11.081928 IP 192.168.116.128.80 > 192.168.136.1.55684: S 4193259095:4193259095(0) ack
685865803 win 5840 <mss 1460,nop,nop,sackOK,nop,wscale 6>
09:35:11.081943 IP 192.168.136.1.55684 > 192.168.116.128.80: . ack 1 win 4380
09:35:11.081955 IP 192.168.136.1.55684 > 192.168.116.128.80: P 1:299(298) ack 1 win 4380
09:35:11.083765 IP 192.168.116.128.80 > 192.168.136.1.55684: . ack 299 win 108 Client B - Src IP 192.168.116.1 - Virtual Server 192.168.136.100:
Clientside:
09:36:11.244040 IP 192.168.116.1.55769 > 192.168.136.100.80: S 3320618938:3320618938(0) win 8192
<mss 1460,nop,wscale 2,nop,nop,sackOK>
09:36:11.244152 IP 192.168.136.100.80 > 192.168.116.1.55769: S 3878120666:3878120666(0) ack
3320618939 win 4380 <mss 1460,nop,wscale 0,sackOK,eol>
09:36:11.244839 IP 192.168.116.1.55769 > 192.168.136.100.80: . ack 1 win 16425
09:36:11.245830 IP 192.168.116.1.55769 > 192.168.136.100.80: P 1:299(298) ack 1 win 16425
09:36:11.245922 IP 192.168.136.100.80 > 192.168.116.1.55769: . ack 299 win 4678 Serverside:
09:36:11.245940 IP 192.168.136.1.55684 > 192.168.116.128.80: P 599:897(298) ack 4525 win 8904
09:36:11.247847 IP 192.168.116.128.80 > 192.168.136.1.55684: P 4525:5001(476) ack 897 win 142 Why was the second client flow permitted by the web server?
A. The idle TCP session from the first client was re-used.
B. A global SNAT is defined.
C. SNAT automap was enabled on the virtual server.
D. A source address persistence profile is assigned to the virtual server.
Answer: A