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UkoKoshka [18]
3 years ago
12

Carlos is using the software development life cycle to create a new app. He has finished coding and is ready to see the output i

t produces. Which stage of the software development life cycle is Carlos ready for next? Coding Design Maintenance Testing
Computers and Technology
2 answers:
dimulka [17.4K]3 years ago
8 0

Answer:

Testing                                   I took the test I definetly testing <3

Nitella [24]3 years ago
4 0

Answer:

Testing

Explanation:

From the question, we understand that Carlos just finished the coding of the app.

In software development life cycle, the coding phase is where Carlos is expected to make use of his choice of programming language to design the app;

This stage is an integral part of the implementation process and according to the question, the coding has been completed;

The next phase or stage after the implementation phase is testing.

Hence, Carlos is getting ready to test the app.

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valentinak56 [21]

Answer:

return values.remove() + values.remove();

Explanation:

The appropriate expression to complete the method, which is designed to return the sum of the two smallest values in the parameter array number is indicated below in bold font :

public static int

sumTwoLowestElements(int[] numbers)

{

PriorityQueue values = new PriorityQueue<>();

for (int num: numbers)

{ values.add(num);

}

return values.remove() + values.remove();  

}

The return statementin programming is often used when a function is ready to return a value to its caller.

6 0
3 years ago
in what way do rules and laws created to address public problems affect individuals groups and business
Anika [276]

Answer:

Explanation:

bussnise

4 0
3 years ago
What should you do before you share your information on the internet?
satela [25.4K]

You have to think about, what could happen to you if you shared this information.

Hope this helped!

~Izzy

4 0
3 years ago
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
A user has created a complex spreadsheet on her workstation containing many graphs and charts. She sent the document to an older
Hatshy [7]

Answer:

The answer to this question can be given as:

To avoid this type of printing problem she must be Install some extra memory in the printer.  

Explanation:

If we want to print only part of a page. So, We use a printer to print. There are many types of printer but we use only laser printer. If there is only one page to print so, we need to add memory in the printer. When a complex graphical document is printed. It will not print the data on the page. In this case, We must update the printer drivers. So, It will automatically fix the problem and the speed of the network connection not harm the quality of the printer output.

5 0
3 years ago
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