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NEW QUESTION: 1
Which of the following is NOT a typical key challenge for testing in a RAD based development approach? 1 credit
A. Time-boxing
B. Re-usable test scripts for (automated) regression testing
C. No complete requirements specification
D. Project management and control
Answer: D
Explanation:
Topic 19, Scenario 11 "Incident Management"
The following is the current incident handling process in used at the company.
Step 1: Incident is documented in the incident Tile with the following information:
- Software module or area where the fault occurred
- Who has reported the fault
- Hardware configuration used for the test that found the fault
- The sequential incident number (1 greater than the last one recorded) Step 2: Developer assigned to fix the fault Step 3: Developer fixes the fault
Step 4: Developer signs off the incident as closed, and it is then removed from the incident file

NEW QUESTION: 2
Proper application of ________ will control but not eliminate microbes that cause MIC so that continued treatment is necessary.
A. None of the above
B. Caustic
C. Ozone
D. Biocides
Answer: D

NEW QUESTION: 3
Given the network diagram.

Which address would successfully summarize only the networks seen?
A. 192.168.16.0/21
B. 192.168.8.0/21
C. These networks cannot be summarized.
D. 192.168.8.0/20
E. 192.168.12.0/20
F. 192.168.0.0/24
Answer: B
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

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