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worty [1.4K]
3 years ago
14

Which of the following is not an operating system?

Computers and Technology
2 answers:
kykrilka [37]3 years ago
8 0
<span>D: Internet Explorer</span>
sergij07 [2.7K]3 years ago
4 0
Internet explorer is a web browser
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Jeff is a financial assistant. He needs to create a document for his client that tracks her stocks and calculates the loss or ga
seropon [69]
Spreadsheet. Brainliest answer plz??
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3 years ago
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Which tool can you use to display hardware utilization statistics that tell you about the operation of your computer?
defon

Performance monitor tool can you use to display hardware utilization statistics.

<h3>what is a performance Monitor?</h3>
  • Performance Monitor is a system monitoring program introduced in Windows NT 3.1. It monitors various activities on a computer such as CPU or memory usage.

To learn more about performance monitoring, refer

to brainly.com/question/12960090

#SPJ4

7 0
2 years ago
Matts has finished running some security automation scripts on three newly deployed Linux servers. After applying intrusion dete
gizmo_the_mogwai [7]

Answer: CPU

Explanation:

The management dashboard refers to the tool that's used in the presentation of the vital k management KPIs in a single place, which is efficiently managed in order to make faster and better decisions.

Based on the information given, after the application of intrusion detection, virus, and malware protection on the Linux images, he will notices an increase in CPU on his server management dashboard.

Therefore, the correct option is C.

5 0
3 years ago
I have six nuts and six bolts. Exactly one nut goes with each bolt. The nuts are all different sizes, but it’s hard to compare t
juin [17]

Answer:

Explanation:

In order to arrange the corresponding nuts and bolts in order using quicksort algorithm, we need to first create two arrays , one for nuts and another for bolts namely nutsArr[6] and boltsArr[6]. Now, using one of the bolts as pivot, we can rearrange the nuts in the nuts array such that the nuts on left side of the element chosen (i.e, the ith element indexed as nutArr[i]) are smaller than the nut at ith position and nuts to the right side of nutsArr[i] are larger than the nut at position "I". We implement this strategy recursively to sort the nuts array. The reason that we need to use bolts for sorting nuts is that nuts are not comparable among themselves and bolts are not comparable among themselves(as mentioned in the question)

The pseudocode for the given problem goes as follows:

// method to quick sort the elements in the two arrays

quickSort(nutsArr[start...end], boltsArr[start...end]): if start < end: // choose a nut from nutsArr at random randElement = nutsArr[random(start, end+1)] // partition the boltsArr using the randElement random pivot pivot = partition(boltsArr[start...end], randElement) // partition nutsArr around the bolt at the pivot position partition(nutsArr[start...end], boltsArr[pivot]) // call quickSort by passing first partition quickSort(nutsArr[start...pivot-1], boltsArr[start...pivot-1]) // call quickSort by passing second partition quickSort(nutsArr[pivot+1...end], boltsArr[pivot+1...end])

// method to partition the array passed as parameter, it also takes pivot as parameter

partition(character array, integer start, integer end, character pivot)

{

       integer i = start;

loop from j = start to j < end

       {

check if array[j] < pivot

{

swap (array[i],array[j])

               increase i by 1;

           }

 else check if array[j] = pivot

{

               swap (array[end],array[j])

               decrease i by 1;

           }

       }

swap (array[i] , array[end])

       return partition index i;

}

7 0
3 years ago
Given an int variable k that has already been declared, use a while loop to print a single line consisting of 88 asterisks. Use
Grace [21]

Answer:

True

Explanation:

The while loop is going to be executed until the condition is false.

Since <em>k</em> is initially equal to 1, the loop will execute 88 times. One asterisk will be printed and <em>k</em> will be incremented by one during each iteration.

When <em>k</em> becomes 89, the condition will be false (89 is not smaller or equal to 88) and the loop will stop.

6 0
3 years ago
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