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Vlad [161]
2 years ago
9

Gray London is a retired race car driver who helped Dale Earnhardt, Jr. get his start. He is writing a book and making a video a

bout the early days or Dale Earnhardt. He is trying to decide whether to market these items directly over the Internet or to use intermediaries. To make this decision, he needs to know the pros and cons of each route. Provide that information and make a recommendation to him.
Computers and Technology
1 answer:
xeze [42]2 years ago
3 0

Answer:

Explanation:

Marketing your product directly over the internet can lead to much greater profits and there are many options that you can choose from in order to target the correct audience. Unfortunately, doing so does require marketing knowledge and is basically like growing a business. If you were to use intermediaries they already have the knowledge necessary to market your product but will take a percentage of the profits, which will ultimately limit your gains. Since Gray London is a race car driver, I am assumming that he does not have any prior technological expertise or marketing expertise, therefore I would recommend using intermediaries.

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no

Explanation:

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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.

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