The solution that would help the client pool their data together is a data integration solution. This type of solution allows for the collection and consolidation of data from various sources into a single, unified database. This can be achieved through a variety of methods, including Extract, Transform, Load (ETL) processes, data warehousing, and master data management. By implementing a data integration solution, the client will be able to access and analyze their data more efficiently, leading to better insights and informed decision-making.
A code in C++ follows:
Explanation:
#include<iostream>
using namespace std;
int calculate_area(int width, int length)
{
int area;
area=width*length;
return area;
}
int main()
{
int a,b,ar,c,d,ar1,n;
cout<<"\nEnter length of room:";
cin>>a;
cout<<"\nEnter width of room:";
cin>>b;
ar=calculate_area(a,b);
cout<<"\nArea of room:"<<ar<<" units";
cout<<"\nEnter length of mat:";
cin>>c;
cout<<"\nEnter width of mat:";
cin>>d;
ar1=calculate_area(c,d);
n=ar/ar1;
cout<<"\nNumber of mats required are:"<<n;
return 0;
}
I think that 3 would be the correct answer
The technology that could be partnered with Cloud to allow the international fast-food chain to gain insights into customer data, including location and order patterns, is called data analytics or big data analytics. It involves using software algorithms and tools to analyze large and complex datasets like customer data, and extract useful insights and patterns that can be used to make informed decisions about customer-focused experiences such as promotions and unique user experiences. Data analytics can help businesses improve customer satisfaction, increase sales and revenue, and gain a competitive advantage in the marketplace.
Question:
When multiple team members are working on a related feature, how often should they integrate their work
Select only one answer.
A) Do the integration midway through the iteration.
B) After they reach a logical end of creating the functionality.
C) In a scheduled daily (or multiple times in a day) frequency.
D) In a scheduled weekly (or multiple times in a week) frequency.
The best option is D.
Explanation:
Multiple integrations per day may become too cumbersome. Leaving the various parts of the project until a logical end for a functionality has been reached may create room for a lot of errors thus resulting in time wasted.
To ensure regular course correction (if required), twice a week would suffice.
Cheers!
Data Security
Yes, Bob is correct that only 8 months is required to pay off the amount of $200 for the ticket.
Given that,
The charge for the plane ticket for the last month is $200.The minimum balance required is $25.And, the number of months given is 8.Based on the above information, the calculation is as follows:
= $25 × 8 months
= $200
Therefore we can conclude that yes, Bob is correct that only 8 months is required to pay off the amount of $200 for the ticket.
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(1)
public int getPlayer2Move(int round)
{
int result = 0;
//If round is divided by 3
if(round%3 == 0) {
result= 3;
}
//if round is not divided by 3 and is divided by 2
else if(round%3 != 0 && round%2 == 0) {
result = 2;
}
//if round is not divided by 3 or 2
else {
result = 1;
}
return result;
}
(2)
public void playGame()
{
//Initializing player 1 coins
int player1Coins = startingCoins;
//Initializing player 2 coins
int player2Coins = startingCoins;
for ( int round = 1 ; round <= maxRounds ; round++) {
//if the player 1 or player 2 coins are less than 3
if(player1Coins < 3 || player2Coins < 3) {
break;
}
//The number of coins player 1 spends
int player1Spends = getPlayer1Move();
//The number of coins player 2 spends
int player2Spends = getPlayer2Move(round);
//Remaining coins of player 1
player1Coins -= player1Spends;
//Remaining coins of player 2
player2Coins -= player2Spends;
//If player 2 spends the same number of coins as player 2 spends
if ( player1Spends == player2Spends) {
player2Coins += 1;
continue;
}
//positive difference between the number of coins spent by the two players
int difference = Math.abs(player1Spends - player2Spends) ;
//if difference is 1
if( difference == 1) {
player2Coins += 1;
continue;
}
//If difference is 2
if(difference == 2) {
player1Coins += 2;
continue;
}
}
// At the end of the game
//If player 1 coins is equal to player two coins
if(player1Coins == player2Coins) {
System.out.println("tie game");
}
//If player 1 coins are greater than player 2 coins
else if(player1Coins > player2Coins) {
System.out.println("player 1 wins");
}
//If player 2 coins is grater than player 2 coins
else if(player1Coins < player2Coins) {
System.out.println("player 2 wins");
}
}
It will provide an instant answer!