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alina1380 [7]
2 years ago
5

A set of activities that performs across the organization creating as output of values to the customer

Computers and Technology
1 answer:
pickupchik [31]2 years ago
4 0
The correct answer is A : Business Process
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During which geologic era was nearly all of Earth's land concentrated into one giant mass called Pangaea?
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One problem with _______ is that often many copies of the same document are made. <br><br>HELPPP ​
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computer or internet is that often many copies of the same document .........

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If I asked you to design a client server application that would work as a progressive slot machine (if you don't know what that
Reptile [31]

Answer:

Explanation:

Iterative vs Concurrent:

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Some idea on on time synchronization is Synchronization of Clocks: Software-Based Solutions

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4.Cristian’s algorithm

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7 0
4 years ago
You are working for the census bureau and you need to write a program in C to do the following: a) count the number of people in
valina [46]

Answer:

See explaination

Explanation:

#include<iostream>

#include<stdio.h>

#include<conio.h>

#include<math.h>

using namespace std;

void main()

{

int RandomData[]={10,5,20,45,20,66,25,31,20,70}; //Random Data

int infant=0,young=0,middle=0,old=0,older=0; //To store count

int n,freq,count=0,high=0; //to store mode and other

double stdDev,mean,sum=0; //to store stdDev mean and sum

for(int i=0;i<(sizeof(RandomData)/sizeof(RandomData[0]));i++) //loop to count

{

if(RandomData[i]>0&&RandomData[i]<=18)

infant++;

else if(RandomData[i]>18&&RandomData[i]<=29)

young++;

else if(RandomData[i]>29&&RandomData[i]<=50)

middle++;

else if(RandomData[i]>50&&RandomData[i]<=69)

old++;

else if(RandomData[i]>69)

older++;

}

double first=0,sec=0; //to store result of stdDev formula first and second part

for(int i=0;i<(sizeof(RandomData)/sizeof(RandomData[0]));i++) //loop to calculate mean mode and stdDev

{

n=RandomData[i];

count=0;

for(int j=0;j<(sizeof(RandomData)/sizeof(RandomData[0]));j++)

{

if(n==RandomData[j])

count++;

}

if(high<count)

{

freq=RandomData[i];

high=count;

}

first+=(pow((RandomData[i]),2)/(sizeof(RandomData)/sizeof(RandomData[0])));

sec+=RandomData[i]/(sizeof(RandomData)/sizeof(RandomData[0]));

sum+=RandomData[i];

}

mean=sum/(sizeof(RandomData)/sizeof(RandomData[0]));

stdDev=sqrt(first-pow((sec),2));

/*------Printing Result-----*/

cout<<"\nCount:\nInfant: "<<infant<<"\nYoung: "<<young<<"\nMiddle: "<<middle;

cout<<"\nOld: "<<old<<"\nOlder: "<<older;

cout<<"\n\nMean: "<<mean<<"\nMode: "<<freq<<"\nStandard Deviation: "<<stdDev;

/*-----------XXXX-------------*/

getch();

}

8 0
3 years ago
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