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  • Exam Number/Code : Introduction-to-IT
  • Exam Name : WGU Introduction to IT
  • Questions and Answers : 260 Q&As
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NEW QUESTION: 1
Which of the following tasks prepares the technical management plan in planning the technical effort
A. Task 9
B. Task 7
C. Task 10
D. Task 8
Answer: A

NEW QUESTION: 2
Given the network diagram.

Which address would successfully summarize only the networks seen?
A. 192.168.8.0/21
B. 192.168.0.0/24
C. 192.168.12.0/20
D. 192.168.16.0/21
E. These networks cannot be summarized.
F. 192.168.8.0/20
Answer: A
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
1 0 0000 1010
1 1 0000 1011
1 2 0000 1100
1 3 0000 1101
1 4 0000 1110
1 5 0000 1111
I've bolded the bits in common there are 5 so 16+5=21 which is mask and address is
1 92.168.8.0

NEW QUESTION: 3
Of the following antenna connector types, which one is the smallest?
A. Male N Connector
B. N Connector
C. RP-TNC
D. MC Connector
Answer: D

NEW QUESTION: 4
Which security benefit does SNMPv3 offer that SNMPv2 does not offer?
A. SSL encryption
B. RADIUS groups
C. Scopes
D. Data integrity
Answer: A

 

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