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natulia [17]
1 year ago
13

Which application development model approaches software development as a continuous, changing process with never-ending versions

, bug fixes, and enhancements?
Computers and Technology
1 answer:
Masteriza [31]1 year ago
5 0

A software application development model that typically approaches software development as a continuous, changing process that has bug fixes, never-ending versions, and enhancements is Agile.

<h3>What is SDLC?</h3>

SDLC is an abbreviation for software development life cycle and it can be defined as a strategic methodology that defines the key steps, phases, or stages for the design, development and implementation of high quality software programs.

<h3>What is Agile software development?</h3>

In Agile software development, the software development team are generally more focused on the continuous production of working (active) software programs with less effort on documentation while creating bug fixes and never-ending versions.

In this context, we can infer and logically deduce that a software application development model that typically approaches software development as a continuous, changing process that has bug fixes, never-ending versions, and enhancements is Agile.

Read more on software development here: brainly.com/question/26324021

#SPJ1

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write a program that computes an integer's checksum. to compute the checksum, break the integer into its constituent
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Using the knowledge in computational language in python it is possible to write a code that computes an integer's checksum.

<h3>Writting the code:</h3>

<em>def checkSum(n):                    # Function definition</em>

<em>    total = 0                       # total variable which returns sum</em>

<em>    c = 0                           # counter variable which count the even position</em>

<em>    while(n!=0):                    # running a loop while n != 0</em>

<em>        r = int(n%10)               #breaking the number into digits</em>

<em>        c=c+1                       # increase counter</em>

<em>        if c%2==0:                  # if counter is even position then double the number </em>

<em>            r = r*2</em>

<em>            while(r!=0):                            # do the process</em>

<em>                total = total + int(r%10)</em>

<em>                r = int(r/10)</em>

<em>            n=int(n/10)    </em>

<em>        else:                       # if counter is odd position then simple add the digit </em>

<em>            total = total + r       # into total</em>

<em>            n=int(n/10)</em>

<em>    return total                    # in the last return the sum of digits</em>

<em># Driver code</em>

<em>n = int(input("Enter Number:"))         # asking user to enter the number </em>

<em>print(checkSum(n))                      # calling the function</em>

See more about python at brainly.com/question/18502436

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Write a program to implement problem statement below; provide the menu for input N and number of experiment M to calculate avera
zalisa [80]

Answer:

Explanation:

#include<iostream>

#include<ctime>

#include<bits/stdc++.h>

using namespace std;

double calculate(double arr[], int l)

{

double avg=0.0;

int x;

for(x=0;x<l;x++)

{

avg+=arr[x];

}

avg/=l;

return avg;

}

int biggest(int arr[], int n)

{

int x,idx,big=-1;

for(x=0;x<n;x++)

{

if(arr[x]>big)

{

big=arr[x];

idx=x;

}

}

return idx;

}

int main()

{

vector<pair<int,double> >result;

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

int choice;

cin>>choice;

while(choice!=2)

{

int n,m;

cout<<"Enter N"<<endl;

cin>>n;

cout<<"Enter M"<<endl;

cin>>m;

int c=m;

double running_time[c];

while(c>0)

{

int arr[n];

int x;

for(x=0;x<n;x++)

{

arr[x] = rand();

}

clock_t start = clock();

int pos = biggest(arr,n);

clock_t t_end = clock();

c--;

running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;

}

double avg_running_time = calculate(running_time,m);

result.push_back(make_pair(n,avg_running_time));

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

cin>>choice;

}

for(int x=0;x<result.size();x++)

{

cout<<result[x].first<<" "<<result[x].second<<endl;

}

}

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3 years ago
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