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disa [49]
3 years ago
5

I need to change the subject before they get onto me. I am only revealing info today if you are a friend.

Computers and Technology
1 answer:
Luden [163]3 years ago
7 0

Answer:

uuuuummmm i have no idea what your doing but tillingan is a doctor and his website is to sell children over internet

Explanation:

im scared of myself ik

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During a test, it is generally best to: A. answer all questions in order B answer the essay questions first C. answer the least
Alecsey [184]

C. answer the least challenging questions first

because then you get those out of the way and have more time to focus on the harder questions.

5 0
3 years ago
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About n processes are time-sharing the CPU, each requiring T ms of CPU time to complete. The context switching overhead is S ms.
Genrish500 [490]

Answer:

(a) Q = (M-(1-n)S)/n

(b)    

When M = 450,

       Q = 82

       % CPU time wasted = 8.889%

When M = 90,

       Q = 10

       % CPU time wasted = 44.44%

When M = 50,

       Q = 2

       % CPU time wasted = 80%

Explanation:

Given  Data:

n = process

T = ms time

Context switch time overhead = S

and M is total overhead

See attached file for the calculation.

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3 years ago
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AP courses deliver college credits

4 0
3 years ago
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which one of the following will reach every element in the array a? 1. for(int i = 0; i <= a.length; i++) 2. for(int i = 0; i
andrey2020 [161]

Answer:

You can simplify the problem down by recognizing that you just need to keep track of the integers you've seen in array that your given. You also need to account for edge cases for when the array is empty or the value you get would be greater than your max allowed value. Finally, you need to ensure O(n) complexity, you can't keep looping for every value you come across. This is where the boolean array comes in handy. See below -

public static int solution(int[] A)

{

int min = 1;

int max = 100000;

boolean[] vals = new boolean[max+1];

if(A.length == 0)

return min;

//mark the vals array with the integers we have seen in the A[]

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

{

if(A[i] < max + 1)

vals[A[i]] = true;

}

//start at our min val and loop until we come across a value we have not seen in A[]

for (int i = 1; i < max; i++)

{

if(vals[i] && min == i)

min++;

else if(!vals[i])

break;

}

if(min > max)

return max;

return min;

}

4 0
1 year ago
Who tryna trade in gta my psn is xMadOrNawx-_-
barxatty [35]
Ok I’ll add you to day
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