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marta [7]
1 year ago
9

What is the difference between the wiring configurations for a residential, 8-position, 8-contact (8P8C) modular plug and jack i

n the Universal Service Order Codes (USOC) and ANSI/TIA/EIA 570-B
Computers and Technology
1 answer:
Molodets [167]1 year ago
7 0

The key variance between how the two wiring configurations described are in the way that the pair numbers three and four are situated or placed.

<h3>What is Service Order Codes (USOC)?</h3>

The above refers to a system of specifications that was created by Bell Systems company to enable the connection of equipment used in buildings both in homes and in public places.

The ANSI/TIA/EIA 570-B on the other hand is the standard that the cabling for telecommunications must meet.

Hence, the primary variance between how the configuration or settings of the wiring for a residential, 8-position, 8-contact (8P8C) modular plug and jack in the Universal Service Order Codes (USOC) and ANSI/TIA/EIA 570-B is in the way that the pair numbers three and four are situated or placed.

Learn more about wiring at:

brainly.com/question/25922783
#SPJ4

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Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
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Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

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