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Lerok [7]
2 years ago
6

zybooks cis 110 challenge activity 9.8.1 Write a program that takes in a positive integer as input, and outputs a string of 1's

and 0's representing the integer in binary. For an integer x, the algorithm is: As long as x is greater than 0 Output x % 2 (remainder is either 0 or 1) x = x / 2 Note: The above algorithm outputs the 0's and 1's in reverse order. Ex: If the input is 6, the output is 011. Your program should define and call a function: Function Integer To Binary(integer num) returns nothing The function should output 1's and 0's representing the integer in binary (in reverse).
Computers and Technology
1 answer:
irakobra [83]2 years ago
7 0

Answer:

public class IntegerToBinary

{

public static void main(String[] args) {

 integerToBinary(6);

}

public static void integerToBinary(int num){

    while(num > 0){

       System.out.print(num%2);

       num = Math.floorDiv(num, 2);

    }

}

}

Explanation:

*The code is in Java.

Create a function called integerToBinary that takes one parameter, num

Inside the function, create a while loop that iterates while the num is greater than 0. Inside the loop, print the num%2. Then, get the floor division of the num by to and assign it to the num.

Inside the main, call the function with parameter 6.

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Katarina [22]

Answer:

The Tp value 0.03 micro seconds as calculated in the explanation below is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP.

Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

  = 15 bits/sec

10 independent connections are established, during parallel downloads,  and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required  is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

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Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation  of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

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The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

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2 years ago
7.8.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's f
Lilit [14]

Answer:

Here is the CoordTransform() function:              

void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){

xValNew = (xVal + 1) * 2;

yValNew = (yVal + 1) * 2; }

The above method has four parameters xVal  and yVal that are used as input parameters and xValNew yValNew as output parameters. This function returns void and transforms its first two input parameters xVal and yVal into two output parameters xValNew and yValNew according to the formula: new = (old + 1) *

Here new variables are xValNew  and yValNew and old is represented by xVal and yVal.

Here the variables xValNew and yValNew are passed by reference which means any change made to these variables will be reflected in main(). Whereas variables xVal and yVal are passed by value.

Explanation:

Here is the complete program:

#include <iostream>

using namespace std;

void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){

xValNew = (xVal + 1) * 2;

yValNew = (yVal + 1) * 2;}

int main()

{ int xValNew;

int yValNew;

int xValUser;

int yValUser;

cin >> xValUser;

cin >> yValUser;

CoordTransform(xValUser, yValUser, xValNew, yValNew);

cout << "(" << xValUser << ", " << yValUser << ") becomes (" << xValNew << ", " << yValNew << ")" << endl;

return 0; }

The output is given in the attached screenshot   

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