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NEW QUESTION: 1
Which three statements about IP Source Guard are true? (Choose three.)
A. When you first enable IP Source Guard on an interface, you may experiencedisruption in IP traffic, until the hosts on the interface receive a new IP address from a DHCP server
B. IP Source Guard is dependent upon DHCP snooping to build and maintain the IP-MAC address binding table or upon manual maintenance of static IP source entries.
C. By default, IP Source Guard is enabled on all interfaces.
D. IP Source Guard requires that DHCP snooping is disabled.
E. IP Source Guard limits IP traffic on an interface to only those resources that have an IP-MAC address binding table entry or static IP source entry.
Answer: A,B,E
Explanation:
Explanation/Reference:
Explanation:

NEW QUESTION: 2
You are planning a design pattern based on the Lambda architecture as shown in the exhibit.

Which Azure services should you use f2 or the cold path? To answer, drag the appropriate services to the correct layers. Each service may be used once, more than once, or not at all. You may need to drag the split bar between panes or scroll to view content.
NOTE: Each correct selection is worth one point.

Answer:
Explanation:

Explanation

Layer 2: Azure Data Lake Storage Gen2
Layer 3: Azure SQL Data Warehouse
Azure SQL Data Warehouse can be used for batch processing.
Note: Lambda architectures use batch-processing, stream-processing, and a serving layer to minimize the latency involved in querying big data.

References:
https://azure.microsoft.com/en-us/blog/lambda-architecture-using-azure-cosmosdb-faster-performance-low-tco-l
https://docs.microsoft.com/en-us/azure/architecture/data-guide/technology-choices/batch-processing

NEW QUESTION: 3
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. A global SNAT is defined.
B. The idle TCP session from the first client was re-used.
C. SNAT automap was enabled on the virtual server.
D. A source address persistence profile is assigned to the virtual server.
Answer: B

NEW QUESTION: 4
For which condition types can you maintain a condition master record for pricing? There are 2 correct answers to this question.
A. Surcharges - KP01
B. Cash discount - SKTO
C. Freight - KF00
D. Costs - VPRS
Answer: A,C