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iogann1982 [59]
3 years ago
13

Assume you are using the text's array-based queue and have just instantiated a queue of capacity 10. You enqueue 5 elements and

then deque elements. Which indices of the internal array elements hold the remaining elements? a) 0 to 2 b) 7 to 9 c) 2 to 4 d) 1 to 3
Computers and Technology
1 answer:
Anika [276]3 years ago
7 0

Full question:

Assume you are using the text's array-based queue and have just instantiated a queue of capacity 10. You enqueue 5 elements and then deque 2 elements. Which indices of the internal array elements hold the remaining elements? a) 0 to 2 b) 7 to 9 c) 2 to 4 d) 1 to 3

Answer and Explanation:

If you instantiate a capacity of 10 for the queue, you have the queue a capacity of not more than 10 values.

To enqueue means to add an element to the array(if it's not full, n<10)

To dequeue means to delete an element from the array or queue(if n! =0)

From the question, for example:

We create a queue of capacity 10:

Queue q(10);

We add elements/enqueue 5 elements to the queue :

q.queueEnqueue(10);

q.queueEnqueue(5);

q.queueEnqueue(8);

q.queueEnqueue(9);

q.queueEnqueue(2);

If we print this out:

q. queueDisplay()

We would get:

10, 5, 8, 9, 2

We remove elements/dequeue 2 elements from the queue :

q. queuedequeue();

q. queuedequeue();

We print it out:

q. queueDisplay()

8 ,9, 2

We observe that deletion/dequeue starts from the front/first index.

We are left with indices 2, 3, 4 or 2 to 4

Note: the above methods/ functions and objects used are merely for example purposes. The queue uses a floating front design approach.

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Stels [109]

Answer:

#include <stdio.h>

int main()

{

float your_value1, your_value2, your_value3, your_value4, your_value5;

printf("Enter a frequency: ");

scanf_s("%f", &your_value1);//storing initial key frequency in your value 1

 

float r = 2.0 / 12;//typing 2.0 so it is treated as float and not int

your_value2 = your_value1 * r * 1; //initial*r*n

your_value3 = your_value1 * r * 2; //initial*r*n

your_value4 = your_value1 * r * 3; //initial*r*n

your_value5 = your_value1 * r * 4; //initial*r*n

printf("%0.2f %0.2f %0.2f %0.2f %0.2f", your_value1, your_value2, your_value3, your_value4, your_value5);

return 0;

}

Explanation:

The purpose of this exercise is to make you understand the difference between float and int. float variables are used when you need decimals in your calculations. int is used when you need integers. The problem in this exercise was the formulation of r. Now r is = 2/12, this means that when we type r as that, the computer assumes that it is an integer and treats it as such. So, it will convert the 0.166667 into 0. To overcome this, all you have to do is type 2.0 instead of 2 alone.

The %0.2 command restricts the float variable to 2 decimal places. By default, it has 6 decimal places.

I have used the function scanf_s instead of scanf simply because my compiler does not work with scanf.

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Answer:

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Explanation:

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Answer:
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Hope this helps!
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Answer:

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