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alexandr402 [8]
2 years ago
7

Smallest kind of computer

Computers and Technology
2 answers:
allochka39001 [22]2 years ago
4 0
As of 2015, the smallest computer is just one cubic centimeter and it’s called the Michigan Micro Mote.
jeka57 [31]2 years ago
3 0
I don’t know if i’m right but like Michigan Micro Mote ?
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Dyn is a cloud-based internet performance management company that provides DNS services for internet websites. It was attacked w
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Answer: ......

Explanation: yes

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Why would a store owner want to calculate a mean
chubhunter [2.5K]

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a store owner would calculate a mean to see the how much a person spends

Explanation:

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As text is typed in the _____ text box, a drop-down list displays options for completing the action, displaying information on t
pickupchik [31]

Answer:

Excel Help

Explanation:

Press F1 key to open the excel help window

Or you can also click the excel help button to launch the help window.

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3 years ago
Recall that within the ArrayBoundedQueue the front variable and the rear variable hold the indices of the elements array where t
Citrus2011 [14]

Answer:

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Explanation:

Options are not made available; However, I'll answer the question base on the following assumptions.

Assumptions

Array name = elements

The front and the rear variable hold the current indices elements

Element to enqueue is "element" without the quotes

To enqueue means to add an element to an already existing queue or to a new queue.

But first, the queue needs to be checked if it can accept a new element or not; in other words, if it's full or not

To do this, we check: if rear < the length of the queue

If this statement is true then it means the array can accept a new element; the new element will be stored at elements[rear] and the rear will be icremented by 1 rear by 1

Else if the statement is false, then an overflow is said to have occurred and the element will not be added.

Going by the above explanation, we have

int n = elements.length;

if(rear < n) {

rear = (rear + 1) % n;

elements[rear] = element;

rear = rear + 1;

}

Additional explanation:

The first line calculates the length of the queue

The if condition on line 2 tests if the array can still accept an element or not

Let's assume this statement is true, then we move to liine 3

Line 3 determines the new position of rear.

Let's assume that n = 6 and current rear is 4.

Line 3 will produce the following result;

rear = (rear + 1) % n;

rear = (4 + 1)% 6

rear = 5%6

rear = 5.

So, the new element will be added at the 5th index

Then line 4 will be equivalent to:

elements[rear] = element;

elements[5] = element;

Meaning that the new element will be enqueued at the 5th index.

Lastly, rear is incremented by 1 to denote the new position where new element can be added.

6 0
3 years ago
What would be the result after the following code is executed? int[] numbers = {40, 3, 5, 7, 8, 12, 10}; int value = numbers[0];
maw [93]

Answer:

The value variable will contain the lowest value in the numbers array.

Explanation:

Given

The given code segment

Required

The result of the code when executed

The illustration of the code is to determine the smallest of the array.

This is shown below

First, the value variable is initialized to the first index element

int value = numbers[0];

This iterates through the elements of the array starting from the second

 for (int i = 1; i < numbers.length; i++) {

This checks if current element is less than value.

     if (numbers[i] < value)

If yes, value is set to numbers[i]; which is smaller than value

value = numbers[i];

<em>Hence, the end result will save the smallest in value</em>

7 0
3 years ago
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