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aalyn [17]
3 years ago
11

Match each item with a statement below. a device used in mobile devices to sense the physical position of the device an area whe

re access to Wi-Fi Internet connectivity is made available, such as at a coffee shop, retail store, or airport a device that contains a disc that is free to move and can respond to gravity as the device is moved based on Mac OS X and is currently used on the iPhone, iPad, and iPod touch by Apple a feature that uses Bluetooth to detect nearby compatible devices, then creates a peer-to-peer network using a Wi-Fi signal between the devices allows a task to be started on a device, such as an iPad, and then pick up that task on another device, such as a Mac desktop or laptop a unique number that identifies a cellular subscription for a device or subscriber, along with its home country and mobile network a unique number that identifies each mobile phone or tablet device worldwide. the process of connecting with another Bluetooth device:__________
Computers and Technology
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

The process of connecting with another Bluetooth device is called <u>pairing</u>

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N=5  key task decision start and end

Explanation:

8 0
3 years ago
7.2 code practice edhesive
goblinko [34]

Answer:

Explanation:

got a 100

7 0
3 years ago
Read 2 more answers
Write a C++ function with the following signature: void readAndConvert() The function takes no parameters and returns no value.
shepuryov [24]

Answer:

The function is as follows:

void readAndConvert(){

   int n; string symbol,name;

   cin>>n;

   cin>>symbol;

   cin.ignore();

   getline (cin,name);

   vector<string> trades;

   string trade;

   for (int inps = 0; inps < n; inps++){

       getline (cin,trade);

       trades.push_back(trade);}

   

   cout<<name<<" ("<<symbol<<")"<<endl;

   for (int itr = 0; itr < n; itr++){

       string splittrade[3];        int k = 0;

       for(int j=0;j<trades.at(itr).length();j++){

           splittrade[k] += trades.at(itr)[j];

           if(trades.at(itr)[j] == ' '){

               k++;    }}

cout<<splittrade[2]<<": "<<floor(stod(splittrade[1]) * stod(splittrade[0]))<<endl;        }

   }

Explanation:

See attachment for complete program where comments are used to explain each line

Download cpp
4 0
3 years ago
Convert ⅖ pie radian to degree​
aniked [119]

Answer:

Convert to a decimal.

22.9183118°

Explanation:

7 0
3 years ago
The birthday paradox says that the probability that two people in a room will have the same birthday is more than half, provided
poizon [28]

Answer:

The Java code is given below with appropriate comments for explanation

Explanation:

// java code to contradict birth day paradox

import java.util.Random;

public class BirthDayParadox

{

public static void main(String[] args)

{

   Random randNum = new Random();

   int people = 5;

   int[] birth_Day = new int[365+1];

   // setting up birthsdays

   for (int i = 0; i < birth_Day.length; i++)

       birth_Day[i] = i + 1;

 

   int iteration;

   // varying number n

   while (people <= 100)

   {

       System.out.println("Number of people: " + people);

       // creating new birth day array

       int[] newbirth_Day = new int[people];

       int count = 0;

       iteration = 100000;

       while(iteration != 0)

       {

           count = 0;

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               // generating random birth day

               int day = randNum.nextInt(365);

               newbirth_Day[i] = birth_Day[day];

           }

           // check for same birthdays

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               int bday = newbirth_Day[i];

               for (int j = i+1; j < newbirth_Day.length; j++)

               {

                   if (bday == newbirth_Day[j])

                   {

                       count++;

                       break;

                   }

               }

           }

           iteration = iteration - 1;

       }

       System.out.println("Probability: " + count + "/" + 100000);

       System.out.println();

       people += 5;

   }

}

}

4 0
4 years ago
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