The transmission control protocol (TCP) layer helps computers to communicate in which of the following ways?

How would you reduce or minimize the size of a "file"?

beks73 [17]

The answer is D Compress

I hope this helps

6 0

3 years ago

Create a game that rolls two dies (number from 1 to 6 on the side) sequentially for 10 times (use loop). If at least once out of

Alex73 [517]

Answer:

Explanation:

The following code is written in Java and loops through 10 times. Each time generating 2 random dice rolls. If the sum is 10 it breaks the loop and outputs a "You Win" statement. Otherwise, it outputs "You Lose"

import java.util.Random;

class Brainly {

public static void main(String[] args) {

UseRandom useRandom = new UseRandom();

boolean youWin = false;

for (int x = 0; x<10; x++) {

int num1 = useRandom.getRandom(6);

int num2 = useRandom.getRandom(6);

if ((num1 + num2) == 10) {

System.out.println("Number 1: " + num1);

System.out.println("Number 2: " + num2);

System.out.println("You Win");

youWin = true;

break;

}

if (youWin == false) {

System.out.println("You Lose");

}

class UseRandom{

public int getRandom(int n)

{

Random r=new Random();

int rand=r.nextInt(n);

return rand;

}}

8 0

3 years ago

rapid prototyping could be an advantageous methodology for developing innovative computer-based instruction. Software engineers

frosja888 [35]

The option that is true for the Student Version above is option d: This is not plagiarism.

<h3>What is plagiarism?</h3>

This is known to be the act of copying other people's work and then taking it as your own.

When you look at the student work, you will see some measures od differences. Hence, The option that is true for the Student Version above is option d: This is not plagiarism.

Learn more about prototyping from

brainly.com/question/14743515
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See full question below

Original Source Material

There is a design methodology called rapid prototyping, which has been used successfully in software engineering. Given similarities between software design and instructional design, we argue that rapid prototyping is a viable method for instructional design, especially for computer-based instruction.

Student Version

Rapid prototyping could be an advantageous methodology for developing innovative computer-based instruction. Software engineers have been successful in designing applications by using rapid prototyping. So it also could be an efficient way to do instructional design.

Which of the following is true for the Student Version above?

a. Word-for-Word plagiarism

b. Paraphrasing plagiarism

c. This is not plagiarism

8 0

2 years ago

Which of these jobs would be most appropriate for someone who majors in information technology? managing a database for a large

Marrrta [24]

Answer:

A good job would be in the IT or getting a job at a network place maybe even a place like Best Buy

Explanation:

7 0

3 years ago

Read 2 more answers

a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv

Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

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3 0

2 years ago