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Dunkirk101
 
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Question for al of you Computer Savvy Geniuses Out There

This is a question pertaining to basic internet protocol

For the slash ( / ) address 255.255.255.252 How many addresses are present?
Old 09-30-2016, 03:29 PM Dunkirk101 is offline  
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growler
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256
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Old 09-30-2016, 05:36 PM growler is offline  
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Dunkirk101
 
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Quote:
Originally Posted by growler View Post
256

Please elaborate on how you achieved this answer
Old 09-30-2016, 05:49 PM Dunkirk101 is offline  
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joemama
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Correct answer in detail:

1. Open nearest window (or don't, makes no diff)
2. Unplug pc (or don't)
3. Toss pc out of previously opened or unopened window
4. Problem solved
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Old 09-30-2016, 06:17 PM joemama is offline  
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Dunkirk101
 
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Quote:
Originally Posted by joemama View Post
Correct answer in detail:

1. Open nearest window (or don't, makes no diff)
2. Unplug pc (or don't)
3. Toss pc out of previously opened or unopened window
4. Problem solved

Ha

I have the answer now...

The slash CIDR or (Classless Inter Domain Routing) number would be indeed 30 (or /30), and withy only two host bits left, the host addresses would be 4

1) address 0 (Network)
2) address 1
3) address 2
4) address 3 (Broadcast)

0 is for Networking and cannot be used
3 is a Broadcast address and cannot be used

This only leaves 2 useable addresses left and would be mainly for used for Point to Point


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Old 10-01-2016, 09:16 AM Dunkirk101 is offline  
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Old 10-03-2016, 12:28 PM Typhoon43 is offline  
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