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Finger [1]
3 years ago
8

Identify the programmer’s responsibility in maximizing the programs reliability by having awareness of the beneficial and harmfu

l impact of the created programs on society, economies, and culture.
Please this is urgent
Computers and Technology
1 answer:
trapecia [35]3 years ago
8 0

Programmers can maximize the reliability of their programs by means of functional analysis.

The form in which programmers can <em>maximize</em> the reliability of their programs in accordance with the awareness of the beneficial and harmful impact on society, economies and cultures is how well they did processes of functional analyses, in which they investigate on market and client needs and derive <em>functional</em> and <em>non-functional</em> parameters, which define the scope and characteristics of the program.

A good functional analysis has an influence of about 80 % on success of programs on fulfilling existing needs and solving existing problems with a <em>reasonable</em> efforts.

We kindly invite to check this question on functional analysis: brainly.com/question/23731043

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What can help establish the focus and organization it relies on? A) Plan B) programming C) organizing D) arranging
serg [7]

Organizational communication helps us to

1) accomplish tasks relating to specific roles and responsibilities of sales, services, and production

2) acclimate to changes through individual and organizational creativity and adaptation

3) complete tasks through the maintenance of policy, procedures, or regulations that support daily and continuous operations

4) develop relationships where “human messages are directed at people within the organization-their attitudes, morale, satisfaction, and fulfillment”

5) coordinate, plan, and control the operations of the organization through management.

So the answer is C.

hope it helps!

7 0
4 years ago
Read 2 more answers
The default setting for a secondary zone's refresh interval is how many minutes
Vinil7 [7]
The default value at which a refresh interval (an interval a sec. server check for zone updates) is 15 minutes.
If this value is increased, the network traffic is reduced. In the eventuality that the refresh interval expires, the secondary zone will contact the primary zone and request it to initiate the zone transfer.
4 0
4 years ago
In which case will the linear search return the lowest value faster than the<br> binary search?
Sveta_85 [38]

Answer:

A linear search is one that scans every record/file until it discovers the value being searched for.

Binary search, on the other hand, is also known as <em>Logarithmic search</em>. It is used to locate the position of a value inside an array that has already been sorted.  

The linear search will return the lowest value faster than the binary search when small arrays are involved.

This will only be feasible when the array is sorted prior.

Cheers!

5 0
3 years ago
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

5 0
3 years ago
two Smallest (10 points). Write a program TwoSmallest.java that takes a set of double command-line arguments and prints the smal
Umnica [9.8K]

Answer:

Below is the program TwoSmallest.java with each step explanation in form of comments.

public class TwoSmallest {                     // class definition

//main class having string args as a parameter

   public static void main(String[] args)

{ if (args.length < 2)

{         //string length must not be less than 2 for proceeding

           System.out.println("Please provide double values as command line arguments");

  }

else {  

// first two entries of string are checked for min1 and min2 and stored in variables with data type double

           double min1 = Double.parseDouble(args[0]), min2 = Double.parseDouble(args[1]), num;

//when min1 will be greater than min2 it will be stored temporary in variable temp having data type double

           if (min1 > min2) {

               double temp = min1;

               min1 = min2;

               min2 = temp;  //value of temp will be stored in min2

           }

//loop will check for each entry remaining until the last character of string

           for (int i = 2; i < args.length; i++) {

               num = Double.parseDouble(args[i]);

               if (num < min1) {

                   min2 = min1;

                   min1 = num;

               } else if (num < min2) {

                   min2 = num;

               }

           }

//min1 will give the 1st minimum number and min2 will give 2nd minimum.

           System.out.println(min1);

           System.out.println(min2);  

//both characters will be printed in sequence

       }

   }

}

4 0
4 years ago
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