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ollegr [7]
2 years ago
12

Which of the following barriers to oral communication is not the fault of the sender or receiver? being unprepared noise not pay

ing attention laziness
Computers and Technology
2 answers:
vredina [299]2 years ago
6 0
All of the given choices are indeed common barriers to oral communication. But among them, "noise" is the only one which may not be the fault of the sender of the receiver. Noise can be generated by nearby sources independent of the sender or receiver.
krek1111 [17]2 years ago
6 0

Answer: The answer is noise.

Explanation:

Oral communication refers to communication with spoken words. It means individuals conversing with one another verbally. It can either be formal or informal. In oral communication there is high level of understanding and transparency, flexibility and the feedback is spontaneous. It is the best type of communication to use in the case of problem resolution.

You might be interested in
Given an array A of size N, and a number K. Task is to find out if it is possible to partition the array A into K contiguous sub
Dennis_Churaev [7]

Answer: First lets solve the Prerequisite part

Lets say we have an input array of N numbers {3,2,5,0,5}. We have to  find number of ways to divide this array into 3 contiguous parts having equal sum. So the output for the above input array will be 2 as there are 2 ways to divide the array. One is (3,2),(5),(0,5) and the other is (3,2),(5,0),(5).

Following are the steps to achieve the above outcome.

  • Let p and q point to the index of array such that sum of array elements from 0 to p-1 is equal to sum of array elements from p to q which is equal to the sum of array elements from q+1 to N-1.  
  • If we see the array we can tell that the sum of 3 contiguous parts is 5. So the condition would be that sum of all array elements should be equal to 5 or sum of each contiguous part is equal to sum of all array elements divided by 5.
  • Now create 2 arrays prefix and postfix of size of input array. Index p of prefix array carries sum of input array elements from index 0 to index p. Index q of postfix array carries sum of input array elements from index p to index N-1
  • Next move through prefix array suppose at the index p of prefix array : value of prefix array == (sum of all input array elements)/5.
  • Search the postfix array for p index found above. Search it starting from p+2 index. Increment the count variable by 1 when the value of postfix array =(sum of all input array elements)/5 and push that index of postfix array into a new array. Use searching algorithm on new array to calculate number of values in postfix array.

Now lets solve the main task

We have an array A of size N and a number K. where A[]= {1,6,3,4,7} N=5 and K=3. We have to find if its possible to partition A into 3 contiguous subarrays such that sum of elements in each subarray is the same. It is possible in this example. Here we have 3 partitions (1,6),(3,4),(7) and sum of each of subarrays is same (1+6) (3+4) (7) which is 7.

Following are the steps to achieve the above outcome.

  • In order create K contiguous subarrays where each subarray has equal sum, first check the condition that sum of all elements in the given array should be divisible by K. Lets name another array as arrsum that will be the size of array A. Traverse A from first to  last index and keep adding current element of A with previous value in arrsum. Example A contains (1,6,3,4,7} and arrsum has {1,7,10,14,21}
  • If the above condition holds, now check the condition that each subarray or partition has equal sum. Suppose we represent sum1 to sum of all element in given array and sum2 of sum of each partition then: sum2 = sum2 / K.
  • Compare arrsum to subarray, begining from index 0 and when it becomes equal to sum2 this means that end of one subarray is reached. Lets say index q is pointing to that subarray.
  • Now from q+1 index find p index in which following condition holds: (arrsum[p] - arrsum[q])=sum2
  • Continue the above step untill K contigous subarrays are found. This loop will break if, at some index, sum2 of any subarray gets greater than required sum2 (as we know that every contiguous subarray should contain equal sum).

A easier function Partition for this task:

int Partition(int A[], int N, int k) // A arra y of size N and number k

{      int sum = 0;    int count = 0;  //variables initialization    

   for(int j = 0; j < N; j++)  //Loop that calculates sum of A

  sum = sum + A[j];        

  if(sum % k != 0) //checks condition that sum of all elements of A should be //divisible by k

   return 0;        

   sum = sum / k;  

   int sum2 = 0;  //represents sum of subarray

  for(int j = 0; j < N; j++) // Loop on subarrays

  {      sum2=sum2 + A[j];  

   if(sum2 == sum)    { //these lines locates subarrays and sum of elements //of subarrays should be equal

       sum2 = 0;  

       count++;  }  }  

/*calculate count of subarrays whose

sum is equal to (sum of complete array/ k.)

if count == k print Yes else print No*/

if(count == k)    

return 1;  

   else

   return 0;  }

6 0
3 years ago
QUICKLY PLEASE!!!
Elza [17]

Answer: As the conventional etiquette, which lays out rules of ethics in social contexts, the purpose of netiquette is to help create and sustain a friendly, relaxed and productive atmosphere for online contact, as well as to avoid putting pressure on the system and creating tension between users.

Explanation:

7 0
3 years ago
For whover needed pont
podryga [215]

Answer:

Thank you have a nice day:)

8 0
2 years ago
Read 2 more answers
Write a program that reads in an integer value for n and then sums the integers from n to 2 * n if n is nonnegative, or from 2 *
devlian [24]

Answer:

Following are the program in c++

First code when using for loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

for(i=n; i<=(2*n); i++)

{

sum1+= i;

}

}

else //  check if n  is negative

{

for(i=(2*n); i<=n; i++)

{

sum1+= i;

}

}

cout<<"The sum is:"<<sum1;

return 0;

Output

Enter the number: 1

The sum is:3

second code when using while loop

#include<iostream> // header file  

using namespace std; // namespace

int main() // main method

{

int n, i, sum1 = 0; // variable declaration

cout<<"Enter the number: "; // taking the value of n  

cin>>n;

if(n > 0) // check if n is nonnegative

{

int i=n;  

while(i<=(2*n))

{

sum1+= i;

i++;

}

}

else //  check if n  is negative

{

int i=n;  

while(i<=(2*n))

{

sum1+= i;

i++;

}

}

cout<<"The sum is:"<<sum1;

return 0;

}

Output

Enter the number: 1

The sum is:3

Explanation:

Here we making 2 program by using different loops but using same logic

Taking a user input in "n" variable.

if n is nonnegative then we iterate the loops  n to 2 * n and storing the sum in "sum1" variable.

Otherwise iterate a loop  2 * n to n and storing the sum in "sum1" variable.  Finally display  sum1 variable .

4 0
3 years ago
so you just gonna bring me a birthday gift on my birthday to my birthday party on my birthday with a birthday gift
Marina86 [1]

Answer:

Happy birthday.?

Explanation:

*GLASS BREAKS*

5 0
2 years ago
Read 2 more answers
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