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erma4kov [3.2K]
3 years ago
6

The difference between a want and a need is a want is not necessary for survival. Things necessary for survival are known as ___

____.
A. needs
B. wants
C. desires
Computers and Technology
1 answer:
Zarrin [17]3 years ago
8 0
A.Needs since they are necessary and needed for survival 
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Please help me with this coding problem :)
leonid [27]

Answer:

ll and lll

Explanation:

list dont be a looser

8 0
2 years ago
Define a function UpdateTimeWindow() with parameters timeStart, timeEnd, and offsetAmount. Each parameter is of type int. The fu
Evgesh-ka [11]

Answer:

C code is given below

Explanation:

// Define a function UpdateTimeWindow() with parameters timeStart, timeEnd, and offSetAmount. Each parameter is of type int. The function adds offSetAmount to each of the first two parameters. Make the first two parameters pass-by-pointer. Sample output for the given program:

#include <stdio.h>

// Define void UpdateTimeWindow(...)

void UpdateTimeWindow(int*timeStart, int* timeEnd, int offSetAmount){

*timeStart = *timeStart+ offSetAmount;

*timeEnd = *timeEnd+ offSetAmount;

}

int main(void) {

  int timeStart = 0;

  int timeEnd = 0;

  int offsetAmount = 0;

  timeStart = 3;

  timeEnd = 7;

  offsetAmount = 2;

  printf("timeStart = %d, timeEnd = %d\n", timeStart, timeEnd);

  UpdateTimeWindow(&timeStart, &timeEnd, offsetAmount);

  printf("timeStart = %d, timeEnd = %d\n", timeStart, timeEnd);

  return 0;

}

5 0
3 years ago
Which of the following is NOT a possible combination of values of the variables in this program when it finishes running?
Nady [450]

Answer:

Definitely D

Explanation:

When you flip a coin it’s a 50/50 chance. Meaning it’s equal.

7 0
3 years ago
Read 2 more answers
Given an array as follows
slava [35]

Answer:

1) Method calcTotal:

  1. public static long calcTotal(long [][] arr2D){
  2.        long total = 0;
  3.        
  4.        for(int i = 0; i < arr2D.length; i++ )
  5.        {
  6.            for(int j = 0; j < arr2D[i].length; j++)
  7.            {
  8.                total = total + arr2D[i][j];
  9.            }
  10.        }
  11.        
  12.        return total;
  13.    }

Explanation:

Line 1: Define a public method <em>calcTotal</em> and this method accept a two-dimensional array

Line 2: Declare a variable, total, and initialize it with zero.

Line 4: An outer for-loop to iterate through every rows of the two-dimensional array

Line 6: An inner  for-loop to iterate though every columns within a particular row.

Line 8: Within the inner for-loop, use current row and column index, i and j, to repeatedly extract the value of each element in the array and add it to the variable total.

Line 12: Return the final total of all the element values as an output

Answer:

2) Method calcAverage:

  1. public static double calcAverage(long [][] arr2D){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            for(int j = 0; j < arr2D[i].length; j++)
  8.            {
  9.                total = total + arr2D[i][j];
  10.                count++;
  11.            }
  12.            
  13.        }
  14.        
  15.        double average = total / count;
  16.        
  17.        return average;
  18.    }

Explanation:

The code in method <em>calcAverage</em> is quite similar to method <em>calcTotal</em>. We just need to add a counter and use that counter as a divisor of total values to obtain an average.

Line 4: Declare a variable, count, as an counter and initialize it to zero.

Line 11: Whenever an element of the 2D array is added to the total, the count is incremented by one. By doing so, we can get the total number of elements that exist in the array.

Line 16: Use the count as a divisor to the total to get average

Line 18: Return the average of all the values in the array as an output.

Answer:

3) calcRowAverage:

  1. public static double calcRowAverage(long [][] arr2D, int row){
  2.        double total = 0;
  3.        int count = 0;
  4.        
  5.        for(int i = 0; i < arr2D.length; i++ )
  6.        {
  7.            if(i == row)
  8.            {
  9.                for(int j = 0; j < arr2D[i].length; j++)
  10.                {
  11.                    total = total + arr2D[i][j];
  12.                    count++;
  13.                }
  14.            }
  15.            
  16.        }
  17.        
  18.        double average = total / count;
  19.        
  20.        return average;
  21.    }

Explanation:

By using method <em>calcAverage </em>as a foundation, add one more parameter, row, in the method <em>calcRowAverage</em>. The row number is used as an conditional checking criteria to ensure only that particular row of elements will be summed up and divided by the counter to get an average of that row.

Line 1: Add one more parameter, row,

Line 8-15: Check if current row index, i, is equal to the target row number, proceed to sum up the array element in that particular row and increment the counter.

5 0
3 years ago
Write a Python program to solve the problem described above. Define a function satisfactory_meal(Meal) which takes a single para
kompoz [17]

def dx(fn, x, delta=0.001):

   return (fn(x+delta) - fn(x))/delta

def solve(fn, value, x=0.5, maxtries=1000, maxerr=0.00001):

   for tries in xrange(maxtries):

       err = fn(x) - value

       if abs(err) < maxerr:

           return x

       slope = dx(fn, x)

       x -= err/slope

   raise ValueError('no solution found')

3 0
2 years ago
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