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Kruka [31]
3 years ago
7

Which of the following is not a good technique for managing used oil

Computers and Technology
2 answers:
dexar [7]3 years ago
7 0
<h3>Answer choices are:</h3>

A) Be prepared for oil spills with the proper absorbents

B) Spills in your shop and any releases on pavement or outside should be poured down a drain

C) Have specific, labeled catch pans available for technicians who are collecting oil

D) Do not use oil containers for antifreeze or other non-similar fluids.

<h3>Answer:</h3><h2>B. Spills in your shop and any releases on pavement or outside should be poured down a drain </h2><h3>Explanation:</h3>

Oil is never spilled into drains or onto pavements as it does not dissolve in water, rather it remains existent in the environment causing harm. Water is a polar liquid, which means that its molecules are more inclined to form stronger bonds with each other as compared to bonds with other molecules. Oil, on the other hand, is non-polar, which is why it does not dissolve in water. As a result of which pouring it down, drains will cause the oil to remain existent in the environment. And its properties are harmful to the environment such as the fact that it does not allow plants to absorb water molecules if it attaches to its roots, or how it reduces soil solubility and permeability.

miv72 [106K]3 years ago
5 0
What are the following choices
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3 years ago
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To convert microseconds to nanoseconds: a. Take the reciprocal of the number of microseconds. b. Multiply the number of microsec
stepladder [879]

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Explanation:

In order to convert microseconds to nanoseconds, it should be noted that one should multiply the number of microseconds by 1,000.

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3 years ago
Given an array of values which initially contains 9 8 7 6 5 4 3 2 1, what does the array look like after bubble sort has execute
Phantasy [73]

Answer:

//Here is code in C++.

//include header

#include <bits/stdc++.h>

using namespace std;

// function to print the array

void out_arr(int inp[], int len)

{

for (int i = 0; i < len; i++)

 cout << inp[i] << " ";

cout << endl;

}

// main function

int main()

{

   // decalare and initialize the array

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int len = sizeof(inp)/sizeof(inp[0]);

int x, y;

// perform bubble sort, only 3 pass

for (x = 0; x < 3; x++)  

{

for (y= 0; y< len-x-1; y++)

 if (inp[y] > inp[y+1])

 {

  int t=inp[y];

  inp[y]=inp[y+1];

  inp[y+1]=t;

 }

 

 cout<<"array after "<<x+1<<" pass"<<endl;

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     out_arr(inp, len);

}

return 0;

}

Explanation:

Declare an integer array "inp" and initialize with {9,8,7,6,5,4,3,2,1}. Find the length of the array.In first pass of bubble sort, largest value is placed at the correct position in the array. And in the second, second largest value is placed at its correct position. After the n pass, array will be sorted. In each pass, each element is compared with the next element. if the value is greater than swap them. This continues for all elements in each pass. In the end of a pass a value is placed correctly. So if there is only 3 pass then 3 largest values will be placed at their right place.

Output:

8 7 6 5 4 3 2 1 9                                                                                                              

array after 2 pass                                                                                                            

7 6 5 4 3 2 1 8 9                                                                                                              

array after 3 pass                                                                                                            

6 5 4 3 2 1 7 8 9

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3 years ago
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