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pentagon [3]
2 years ago
15

This can change the fit of your respirator. A) A mustacheB) A beardC) Weight gainD) All of the above

Computers and Technology
1 answer:
nordsb [41]2 years ago
5 0

Answer:

D) All of the above

Explanation:

"Fit testing" is important when it comes to respirators. This ensures the respiratory safety of the wearer. When it comes to tight-fitting respirators, a good fit is very important. This means that the respirator will<em> seal to the skin</em>, so it is important not to have any facial hair like<em> beards, mustache, or stubble</em>. OHSA requires that the person should be<u> clean-shaven</u>.

Fit testing is also done <u><em>annually</em></u> in order to <em>adjust to any physical changes</em> such as <em>weight gain</em> or <em>dental work.</em> So, all of the choices above can change the fit of the respirator.

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Answer:As in merge sort, the time for a given recursive call on an n-element subarray is Θ(n). In merge sort, that was the time for merging, but in quicksort it's the time for partitioning.

Explanation:

Worst-case running time:

When quicksort always has the most unbalanced partitions possible, then the original call takes cncnc, n time for some constant ccc, the recursive call on n-1n−1n, minus, 1 elements takes c(n-1)c(n−1)c, left parenthesis, n, minus, 1, right parenthesis time, the recursive call on n-2n−2n, minus, 2 elements takes c(n-2)c(n−2)c, left parenthesis, n, minus, 2, right parenthesis time, and so on. cn+c(n−1)+c(n−2)+⋯+2c=c(n+(n−1)+(n−2)+⋯+2)

=c((n+1)(n/2)−1)

The last line is because 1 + 2 + 3 +...... n is the arithmetic series

Best-case running time:

Quicksort's best case occurs when the partitions are as evenly balanced as possible: their sizes either are equal or are within 1 of each other. The former case occurs if the subarray has an odd number of elements and the pivot is right in the middle after partitioning, and each partition has (n-1)/ 2 elements. The latter case occurs if the subarray has an even number n of elements and one partition has n/2 elements with the other having n/2-1.In either of these cases, each partition has at most n/2 elements.

Using big-Θ notation, we get the same result as for merge sort: Θ(nlog2n)

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3 years ago
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Answer:

public class Employee {

   private int id;

   private String empName;

   private double salary;

   

   public void displayEmployee(){

       System.out.print("Employee ID: " + id + "\nEmployee name: " + empName + "\nSalary: " + salary);

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}

Explanation:

- Declare the class variables

- Write a method called <em>displayEmployee</em> to display the fields

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Answer:

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