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NEW QUESTION: 1
Given the SAS data set PERM.STUDENTS:
PERM.STUDENTS NAMEAGE ---------------- Alfred14 Alice13 Barbara13
Carol14
The following SAS program is submitted:
libname perm 'SAS data library';
data students;
set perm.students;
file 'file specification';
put name $ age;
<insert statement here>
run;
The following double-spaced file is desired as output
Alfred 14
Alice 13
Barbara 13
Carol 14
Which statement completes the program and creates the desired file?
A. put
B. double;
C. put _null_;
D. put/;
Answer: A

NEW QUESTION: 2
Drag and drop the DHCP messages from the left onto the correct uses on the right.

Answer:
Explanation:

Explanation

DHCPACK
The server-to-client communication with configuration parameters, including committed network address.
DHCPINFORM
The client-to-server communication, asking for only local configuration parameters that the client already has externally configured as an address.
DHCPNAK
The server-to-client communication, refusing the request for configuration parameter.
DHCPDECLINE
The client-to-server communication, indicating that the network address is already in use

NEW QUESTION: 3
Given the network diagram.

Which address would successfully summarize only the networks seen?
A. 192.168.12.0/20
B. 192.168.8.0/20
C. These networks cannot be summarized.
D. 192.168.8.0/21
E. 192.168.16.0/21
F. 192.168.0.0/24
Answer: D
Explanation:
let's suppose it is a /20 then we would have addresses from 192.168.0.1 to 192.168.15.255 Now let's suppose it is a /21 then we would have addresses from 192.168.8.1 to 192.168.15.255 So both summaries encompass the networks we want to summarize but the second one is the most restrictive one as it only encompasses the networks we were asked to summarize and not others so it is the correct summary. In fact just count the number of subnets which is 8 and find the exponent of 2 which is 8, that gives you 3 and to find out the summary mask just do /24 - 3 = /21. if you make it in binary you'll find out the answer too: the interesting octet is 3rd one so let's convert in binary. 8 0000 1000 9 0000 1001 10 0000 1010 11 0000 1011 12 0000 1100 13 0000 1101 14 0000 1110 15 0000 1111 I've bolded the bits in common there are 5 so 16+5=21 which is mask and address is 192.168.8.0

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