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Yuri [45]
3 years ago
11

Which of the following is true about occupations within the STEM fields? Many occupations use mathematics, even if it is not the

primary focus of a given job. Engineering occupations rely solely on technology concepts to solve problems. Scientists assist technicians by collecting information. Most technology occupations require a higher education degree.​
Computers and Technology
2 answers:
Kazeer [188]3 years ago
8 0

Answer:

Many occupations use mathematics, even if it is not the primary focus of a given job.

Explanation:

Think about software developers, they develop apps, websites, and other things, but they also use math in the process. Scientists use statistics. Mechanics use math to make sure their measurements are right. Therefore, I think your best bet would be

A. Many occupations use mathematics, even if it not the primary focus of a given job.

bearhunter [10]3 years ago
3 0

Answer:

Explanation:

Think about software developers, they develop apps, websites, and other things, but they also use math in the process. Scientists use statistics. Mechanics use math to make sure their measurements are right. Therefore, I think your best bet would be

A. Many occupations use mathematics, even if it not the primary focus of a given job.

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Therefore, we would add 1 to an array element whose subscript would be represented by regionNumber since we are interested to know the number of customer owing in each of the 12 sales regions.

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Please select the word from the list that best fits the definition
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d

Explanation:

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Two cars A and B leave an intersection at the same time. Car A travels west at an average speed of x miles per hour and car B tr
drek231 [11]

Answer:

Here is the C++ program:

#include <iostream>  // to use input output functions

#include <cmath>  // to use math functions like sqrt()

#include <iomanip>  //to use setprecision method

using namespace std;   //to access objects like cin cout

int main ()  {  //start of main function

  double speedA;  //double type variable to store average speed of car A

  double speedB;  //double type variable to store average speed of car B

  int hour;  //int type variable to hold hour part of elapsed time

  int minutes;  //int type variable to hold minutes part of elapsed time

  double shortDistance;  // double type variable to store the result of shortest distance between car A and B

  double distanceA;  //stores the distance of carA

  double distanceB;  //stores the distance of carB

  double mins,hours;   //used to convert the elapsed time

cout << "Enter average speed of car A: " << endl;  //prompt user to enter the average speed of car A

cin >> speedA;   //reads the input value of average speed of car A from user

cout << "Enter average speed of car B: " << endl ;  //prompt user to enter the average speed of car B

cin >> speedB;   //reads the input value of average speed of car A from user

cout << "Enter elapsed time (in hours and minutes, separated by a space): " << endl;  //prompts user to enter elapsed time

cin>> hour >> minutes;    //reads elapsed time in hours and minutes

  mins = hour * 60;  //computes the minutes using value of hour

  hours = (minutes+mins)/60;     //computes hours using minutes and mins

distanceA = speedA * (hours);  // computes distance of car A

distanceB = speedB * (hours);   //computes distance of car B

   shortDistance =sqrt((distanceA * distanceA) + (distanceB * distanceB));   //computes shortest distance using formula √[(distanceA)² + (distanceB)²)]

cout << "The (shortest) distance between the cars is: "<<fixed<<setprecision(2)<<shortDistance;

//display the resultant value of shortDistance up to 2 decimal places

Explanation:

I will explain the program with an examples:

Let us suppose that the average speeds of cars are:

speedA = 70

speedB = 55

Elapsed time in hours and minutes:

hour = 2

minutes = 30

After taking these input values the program control moves to the statement:

mins = hour * 60;  

This becomes

mins = 2 * 60

mins = 120

Next

hours = (minutes+mins)/60;

hours = (30 + 120) / 60

         = 150/60

hours = 2.5

Now the next two statements compute distance of the cars:

distanceA = speedA * (hours);  

this becomes

distanceA = 70 * (2.5)

distanceA = 175

distanceB = speedB * (hours);

distanceB = 55 * (2.5)

distanceB = 137.5

Next the shortest distance between car A and car B is computed:

shortDistance = sqrt((distanceA * distanceA) + (distanceB * distanceB));

shortDistance = sqrt((175 * 175) + (137.5 * 137.5))

                        = sqrt(30625 + 18906.25)

                        = sqrt(49531.25)

                        =  222.556173

shortDistance =  222.56

 

Hence the output is:

The (shortest) distance between the cars is: 222.56        

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