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Elis [28]
3 years ago
15

Which statement about synchronous communication is true?

Computers and Technology
1 answer:
tangare [24]3 years ago
3 0
In general, synchronous communication means you have to wait for the answer all the time. The programming logic is simpler, but the cost that you spend a lot of time waiting.

If the options are:

<span>a. The people communicating don't need to be online at the same time.
b. There is lag time in the communication.
c. The communication occurs in real time.

a is false, you do need to be online to receive the message
b is true, typically you continue only after an acknowledgement
c is true, you wait for acknowledgement that occurs in real time (not necessarily fast though)</span>
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Which of the following correctly revises the passive voice of the sentence above to active voice?
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In this problem, there will be the construction of a program that reads in a sequence of integers from standard input until 0 is
Sedaia [141]

Answer:

Check the explanation

Explanation:

package util;

import java.util.ArrayList;

import java.util.Iterator;

import java.util.List;

import java.util.Scanner;

public class Assignment9 {

public Assignment9() {

List<Integer> numList = new ArrayList<Integer>();

Scanner sc = new Scanner(System.in);

System.out.println("Please enter the numbers , press 0 to stop");

int x = 0;

do {

x = sc.nextInt();

numList.add(x);

} while (x != 0);

int size = numList.size();

Integer[] numArray = new Integer[size];

Iterator<Integer> it = numList.iterator();

int i = 0;

while (it.hasNext()) {

numArray[i] = it.next();

i++;

}

int min = findMin(numArray, 0, 2);

System.out.println("The minimum number is " + min);

int sum = computeNegativeSum(numArray, 0, 2);

System.out.println("The sum of the negative numbers is " + sum);

int sumOdd = computeSumAtOdd(numArray, 0, 3);

System.out

.println("The sum of the numbers at odd indexes is " + sumOdd);

int countEven = computeCountEven(numArray, 0, 3);

System.out.println("The total count of even integers is " + countEven);

}

/**

* This method is used to compute the minimum number

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int findMin(Integer[] NumArray, int startIndex, int endIndex) {

if (startIndex == endIndex)// base case50.

{

return NumArray[startIndex]; // return value is there is only one

// entry

} else if (findMin(NumArray, startIndex, endIndex - 1) < NumArray[endIndex]) {

return findMin(NumArray, startIndex, endIndex - 1);

} else {

return NumArray[endIndex];

}

}

/**

* This method is used to find the sum of negative numbers in the array

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int computeNegativeSum(Integer[] NumArray, int startIndex,

int endIndex) {

if (startIndex == endIndex) {

if (NumArray[startIndex] > 0) {

return 0;

} else {

return NumArray[startIndex];

}

} else if (NumArray[endIndex] < 0) {

return computeNegativeSum(NumArray, startIndex, endIndex - 1)

+ NumArray[endIndex];

} else {

return computeNegativeSum(NumArray, startIndex, endIndex - 1); // if

}

}

/**

* This method is used to find the sum of numbers at odd indexes (1,3, 5,...),

*

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int computeSumAtOdd(Integer[] NumArray, int startIndex, int endIndex) {

if (startIndex == endIndex) {

if (startIndex % 2 == 1) {

return NumArray[startIndex];

} else {

return 0;

}

} else {

if (endIndex % 2 == 1) {

return computeSumAtOdd(NumArray, startIndex, endIndex - 1)

+ NumArray[endIndex];

} else {

return computeSumAtOdd(NumArray, startIndex, endIndex - 1);

}

}

}

/**

* This method is used to find the number of even numbers within the array

*

* "at"param NumArray

* "at"param startIndex

* "at"param endIndex

* "at"return

*/

public int computeCountEven(Integer[] NumArray, int startIndex, int endIndex) {

if (startIndex == endIndex) {

if (NumArray[startIndex] % 2 == 0) {

return NumArray[startIndex];

} else {

return 0;

}

} else if (NumArray[endIndex] % 2 == 0) {

return computeCountEven(NumArray, startIndex, endIndex - 1)

+ NumArray[endIndex];

} else {

return computeCountEven(NumArray, startIndex, endIndex - 1); // if

}

}

public static void main(String args[]) {

Assignment9 assignment9 = new Assignment9();

}

}

Sample Output is :

*************************

Please enter the numbers , press 0 to stop

5

7

3

2

4

-7

-3

0

The minimum number is 3

The sum of the negative numbers is 0

The sum of the numbers at odd indexes is 9

The total count of even integers is 2

8 0
3 years ago
CONDUCT INTERVIEWS
iragen [17]

Answer:

you have to interview people around you with the questions that you put that you my have then record your interviews into that box.

Explanation:

...

6 0
3 years ago
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