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Ket [755]
2 years ago
5

Two technicians are discussing lung protection in the shop. Technician A says that respiratory-protection devices can prevent yo

u from inhaling harmful dusts, gases, or vapors. Technician B says that dust masks are made of heavy plastic, metal, and safety glass and offer a higher level of protection than a respirator. Who is correct? A. Technician B only B. Technician A only C. Neither Technician A nor B D. Both Technicians A and B
Computers and Technology
2 answers:
3241004551 [841]2 years ago
7 0

The correct answer is technician A.

A respiratory-protective equipment is used to protect the wearer against inhalation of hazardous substances in the workplace air. It covers the nose and mouth or sometimes the entire face. Respirators may be air purifying, tight fitting, or loose fitting. We should wear respirators only when adequate control of exposure cannot be achieved through other means. A gas mask on the other hand, is made up of light material like paper and does not offer higher level protection than respirator.

Sveta_85 [38]2 years ago
5 0
The correct answer is Technician A ONLY. I had this on my test, so im sure of my answer!
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What is the output of 1101 x 10 == 11000 + 10?
Marta_Voda [28]
== is an operator that returns a boolean if both operand are equal.

1101 * 10 is 11010

11000 + 10 is 11010


11010 == 11010

Thus the output would be True.
5 0
1 year ago
Write a program to declare a matrix A[][] of order (MXN) where ‘M’ is the number of rows and ‘N’ is the
Liula [17]

Answer:

import java.io.*;

import java.util.Arrays;

class Main {

   public static void main(String args[])

   throws IOException{

       // Set up keyboard input

       InputStreamReader in = new InputStreamReader(System.in);

       BufferedReader br = new BufferedReader(in);

 

       // Prompt for dimensions MxN of the matrix

       System.out.print("M = ");

       int m = Integer.parseInt(br.readLine());

       System.out.print("N = ");

       int n = Integer.parseInt(br.readLine());

       // Check if input is within bounds, exit if not

       if(m <= 2 || m >= 10 || n <= 2 || n >= 10){

           System.out.println("Matrix size out of range.");

           return;

       }

       // Declare the matrix as two-dimensional int array

       int a[][] = new int[m][n];

 

       // Prompt for values of the matrix elements

       System.out.println("Enter elements of matrix:");

       for(int i = 0; i < m; i++){

           for(int j = 0; j < n; j++){

               a[i][j] = Integer.parseInt(br.readLine());

           }

       }

       // Output the original matrix

       System.out.println("Original Matrix:");

       printMatrix(a);

       // Sort each row

       for(int i = 0; i < m; i++){

         Arrays.sort(a[i]);

       }

       // Print sorted matrix

       System.out.println("Matrix after sorting rows:");

       printMatrix(a);

   }

   // Print the matrix elements separated by tabs

   public static void printMatrix(int[][] a) {

       for(int i = 0; i < a.length; i++){

           for(int j = 0; j < a[i].length; j++)

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

           System.out.println();

       }

   }

}

Explanation:

I fixed the mistake in the original code and put comments in to describe each section. The mistake was that the entire matrix was sorted, while only the individual rows needed to be sorted. This even simplifies the program. I also factored out a printMatrix() method because it is used twice.

4 0
1 year ago
Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer
NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

6 0
2 years ago
During an election year, why would a senator want to determine a mode?
ale4655 [162]

Answer:

To determine which issue is most important to the general public.

Explanation:

Took the Edge assignment!

3 0
1 year ago
Which type of styles can be applied to a word, phrase, or sentence?
Anna [14]

Answer:

I think the answer is going to be paragraph

6 0
1 year ago
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