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leonid [27]
3 years ago
7

What does “human-like” mean when we are talking about a machine?

Computers and Technology
1 answer:
pav-90 [236]3 years ago
8 0
When talking about a machine and describing it as “human-like” it is suggesting that it had human like characteristics for the inanimate object.
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Should spreadsheets be used to keep an address list for holiday cards?<br> YES<br> ONO<br> Hurry pls
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No

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2 years ago
____ is designed for short-distance connections and has only a few lambdas, with a greater space between lambdas. Group of answe
givi [52]

Answer:

The answer is Coarse wavelength division multiplexing (CWDM)

Explanation:

CWDM is the wavelength designed for short- distance connections  and has only a few lambdas. It is mostly used nowadays as a type of wavelength technologies deployed in cities and inner access ways. It is inexpensive and thus a cost effective wavelength  division multiplexing. Cost effective lasers can be used in CWDM due to the spacing of about 20nm between its wavelengths.

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3 years ago
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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
kow [346]

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

5 0
3 years ago
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