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NEW QUESTION: 1
Which statements correctly describe the features of functions and procedures? (Choose all that apply.)
A. A function can contain zero or more parameters that are transferred from the calling environment.
B. A procedure can be executed as part of a SQL expression or as a PL/SQL statement,
C. A function can return multiple values using a single return clause,
D. A procedure can contain a return statement without a value.
Answer: D
Explanation:
Reference:http://docs.oracle.com/cd/B19306_01/appdev.102/b14261/subprograms.htm(usi ng the return statement)

NEW QUESTION: 2
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 09:36:11.247847 IP 192.168.116.128.80 > 192.168.136.1.55684: P 4525:5001(476) ack 897 win
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. A source address persistence profile is assigned to the virtual server.
D. SNAT automap was enabled on the virtual server.
Answer: A

NEW QUESTION: 3



A. Option E
B. Option A
C. Option B
D. Option C
E. Option D
Answer: E

NEW QUESTION: 4

A. Option A
B. Option B
C. Option C
D. Option D
Answer: D