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masya89 [10]
3 years ago
11

The following code segment is a count-controlled loop going from 1 to 5. At each iteration, the loop counter is either printed o

r put on a queue depending on the result of Boolean function RanFun(). (The behavior of RanFun() is immaterial. At the end of the loop, the items on the queue are dequeued and printed. Because of the logical properties of a queue, this code segment cannot print certain sequences of the values of the loop counter. Is the following output possible using a queue: 1 2 3 4 5?
Computers and Technology
1 answer:
quester [9]3 years ago
4 0

Answer:

True.

Explanation:

For all the count values, if Ranfun() returns true then all the values of count would be printed. Then the sequence of count values printed would be 1 2 3 4 5. Also for all the values of count if Ranfun() returns false then all the values would be enqueued. When we try to dequeue elements from queue we would get them in the same order of their entry into the queue. Thus, the sequence would be 1 2 3 4 5. Hence, we can say that this output sequence is possible. TRUE.

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Which answer choice correctly distinguishes among the three pieces of data?
wolverine [178]

Answer:

a. 1 is a packet, 2 is data, 3 is a frame.

Explanation:

And what is not  mentioned is segment which used TCP/UDP and is part of Transport layer. The packet carries the destination and sender IP address, and is part of the Network Layer. The frame has the Mac address of destination device and senders device and is part of data link layer.

Hence segment has no IP address, hence b. is not correct. Also, data cannot have the IP Address, and Frame has the MAC address, Hence, the above answer. And this arrangement is part of Data Encapsulation.

Also keep in mind data can be anything like a series of bits, or any and it can or not have a header.

7 0
3 years ago
Draw the Abstract Syntax Trees for the following statements and represent them in text form. i) 1+2+3 ii) 6÷3×4+3
MArishka [77]

Answer:

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Explanation:

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5 0
3 years ago
Write a method that prints on the screen a message stating whether 2 circles touch each other, do not touch each other or inters
beks73 [17]

Answer:

The method is as follows:

public static void checkIntersection(double x1, double y1, double r1, double x2, double y2, double r2){

    double d = Math.sqrt(Math.pow((x1 - x2),2) + Math.pow((y1 - y2),2));

    if(d == r1 + r2){

        System.out.print("The circles touch each other");     }

    else if(d > r1 + r2){

        System.out.print("The circles do not touch each other");     }

    else{

        System.out.print("The circles intersect");     }

}

Explanation:

This defines the method

public static void checkIntersection(double x1, double y1, double r1, double x2, double y2, double r2){

This calculate the distance

    double d = Math.sqrt(Math.pow((x1 - x2),2) + Math.pow((y1 - y2),2));

If the distance equals the sum of both radii, then the circles touch one another

<em>     if(d == r1 + r2){</em>

<em>         System.out.print("The circles touch each other");     }</em>

If the distance is greater than the sum of both radii, then the circles do not touch one another

<em>    else if(d > r1 + r2){</em>

<em>         System.out.print("The circles do not touch each other");     }</em>

If the distance is less than the sum of both radii, then the circles intersect

<em>    else{</em>

<em>         System.out.print("The circles intersect");     }</em>

}

6 0
3 years ago
Real GDP removes any inflation from the calculation of GDP. Why is it important to remove a general increase in prices from the
Katen [24]
<span>C. Not including inflation into the caluculation will over-estimate the GDP.</span>
8 0
4 years ago
given an array of integers a, your task is to count the number of pairs i and j (where 0 ≤ i &lt; j &lt; a.length), such that a[
Nimfa-mama [501]

Using the knowledge of computational language in C++ it is possible to write a code that given an array of integers a, your task is to count the number of pairs i and j.

<h3>Writting the code:</h3>

<em>// C++ program for the above approach</em>

<em> </em>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em> </em>

<em>// Function to find the count required pairs</em>

<em>void getPairs(int arr[], int N, int K)</em>

<em>{</em>

<em>    // Stores count of pairs</em>

<em>    int count = 0;</em>

<em> </em>

<em>    // Traverse the array</em>

<em>    for (int i = 0; i < N; i++) {</em>

<em> </em>

<em>        for (int j = i + 1; j < N; j++) {</em>

<em> </em>

<em>            // Check if the condition</em>

<em>            // is satisfied or not</em>

<em>            if (arr[i] > K * arr[j])</em>

<em>                count++;</em>

<em>        }</em>

<em>    }</em>

<em>    cout << count;</em>

<em>}</em>

<em> </em>

<em>// Driver Code</em>

<em>int main()</em>

<em>{</em>

<em>    int arr[] = { 5, 6, 2, 5 };</em>

<em>    int N = sizeof(arr) / sizeof(arr[0]);</em>

<em>    int K = 2;</em>

<em> </em>

<em>    // Function Call</em>

<em>    getPairs(arr, N, K);</em>

<em> </em>

<em>    return 0;</em>

<em>}</em>

See more about C++ code at brainly.com/question/17544466

#SPJ4

4 0
1 year ago
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