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Mamont248 [21]
4 years ago
10

List at least six things you would check for if you were asked to evaluate the workspace of an employee for ergonomics

Computers and Technology
1 answer:
Evgen [1.6K]4 years ago
0 0

Assessing the risk that surrounds stationary work of employees spending hours at their stations is essential. Below is a list of fundamental ergonomic principles that help identify ergonomic risk factors.


<span>1.       </span>Are the employees maintained in a neutral posture?

<span>2.       </span>Does the stationery allow for movement and stretching?

<span>3.       </span>Is there adequate lighting?

<span>4.       </span>Are chairs adequately adjustable?

<span>5.       </span>Are there appropriate foot rest?

<span>6.       </span>Is there extra storage for better desk organization?






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Explanation:

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Write a program in qbasic to accept a character and check it is vowel or consonant​
Alborosie

Answer:

In sub procedure or normal program?

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Take the average of some numbers. Show all the numbers that are below average. You can assume that there will not be more than 2
olchik [2.2K]

Here is code in java.

import java.util.*; // import package

class Main //  creating class

{

public static void main (String[] args) // main method

{

Scanner br=new Scanner(System.in); // scanner class for input

       int []a=new int[20]; // declaring array

       int count=0,x,j; // creating variable

       int sum =0;

       System.out.println("Enter a weight (0 to stop):");

       x=br.nextInt(); // taking input

while(x!=0) // iterating over the loop

       {

           a[count] = x;

           count++;

           System.out.println("Enter a weight (0 to stop):");

           x=br.nextInt();

}

       for( j=0;j<count;j++) // iterating over the loop

       {

       sum+=a[j];

       }

       double avg = sum/(double)count;

       System.out.println("The total is: "+sum);

       System.out.println("The average is: "+avg);

       System.out.println("Here are all the numbers less than the average: ");

     for( j=0;j<count;j++) //  // iterating over the loop

       {

           if(a[j]<avg)

           System.out.print(a[j]+" ");

       }

}

}

Explanation:

First create an object of "Scanner" class to read input from the user.

Create an array of size 20, which store the number given by user.Ask user

to give input and keep the count of number in the variable "count". If user

give input 0 then it will stop taking input.And then calculate sum of all the

input.It will calculate average by dividing the sum with count and print it.

It will compare the all elements of array with average value, if the element

is less than average, it will print that element.

Output:

Enter a weight (0 to stop):

1

Enter a weight (0 to stop):

2

Enter a weight (0 to stop):

3

Enter a weight (0 to stop):

4

Enter a weight (0 to stop):

5

Enter a weight (0 to stop):

6

Enter a weight (0 to stop):

7

Enter a weight (0 to stop):

8

Enter a weight (0 to stop):

9

Enter a weight (0 to stop):

10

Enter a weight (0 to stop):

0

The total is: 55

The average is: 5.5

Here are all the numbers less than the average:

1 2 3 4 5

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Answer:

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UDP (User Datagram Protocol) this is a protocol faster than TCP because this method doesn't establish a connection to sent data, in this case, always sent data, but TCP is more secure than UDP, and for that UDP is used to transfer music or videos, and TCP websites and database.

UDP doesn't need acknowledgment is done by UDP, is only concerned with speed.

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4 years ago
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