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Setler79 [48]
3 years ago
13

Insurance can help you:

Computers and Technology
2 answers:
Fittoniya [83]3 years ago
8 0
If your house is bern insurance can help 
seraphim [82]3 years ago
3 0
Yes insurance can help you
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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
April's manager has asked her to provide the financial service manager with the accounts that have not received payment. What sh
Irina-Kira [14]
C should be the answer to this I hope that this helps
4 0
4 years ago
Read 2 more answers
In this lab you will write a program that simulates a mouse in a maze. The maze will have one exit location. The mouse will star
sergij07 [2.7K]

Answer:

/* C/C++ program to solve Rat in a Maze problem using  

backtracking */

#include <stdio.h>  

// Maze size  

#define N 4  

bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N]);  

/* A utility function to print solution matrix sol[N][N] */

void printSolution(int sol[N][N])  

{  

for (int i = 0; i < N; i++) {  

 for (int j = 0; j < N; j++)  

  printf(" %d ", sol[i][j]);  

 printf("\n");  

}  

}  

/* A utility function to check if x, y is valid index for N*N maze */

bool isSafe(int maze[N][N], int x, int y)  

{  

// if (x, y outside maze) return false  

if (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)  

 return true;  

return false;  

}  

/* This function solves the Maze problem using Backtracking. It mainly  

uses solveMazeUtil() to solve the problem. It returns false if no  

path is possible, otherwise return true and prints the path in the  

form of 1s. Please note that there may be more than one solutions,  

this function prints one of the feasible solutions.*/

bool solveMaze(int maze[N][N])  

{  

int sol[N][N] = { { 0, 0, 0, 0 },  

    { 0, 0, 0, 0 },  

    { 0, 0, 0, 0 },  

    { 0, 0, 0, 0 } };  

if (solveMazeUtil(maze, 0, 0, sol) == false) {  

 printf("Solution doesn't exist");  

 return false;  

}  

printSolution(sol);  

return true;  

}  

/* A recursive utility function to solve Maze problem */

bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N])  

{  

// if (x, y is goal) return true  

if (x == N - 1 && y == N - 1) {  

 sol[x][y] = 1;  

 return true;  

}  

// Check if maze[x][y] is valid  

if (isSafe(maze, x, y) == true) {  

 // mark x, y as part of solution path  

 sol[x][y] = 1;  

 /* Move forward in x direction */

 if (solveMazeUtil(maze, x + 1, y, sol) == true)  

  return true;  

 /* If moving in x direction doesn't give solution then  

 Move down in y direction */

 if (solveMazeUtil(maze, x, y + 1, sol) == true)  

  return true;  

 /* If none of the above movements work then BACKTRACK:  

  unmark x, y as part of solution path */

 sol[x][y] = 0;  

 return false;  

}  

return false;  

}  

// driver program to test above function  

int main()  

{  

int maze[N][N] = { { 1, 0, 0, 0 },  

    { 1, 1, 0, 1 },  

    { 0, 1, 0, 0 },  

    { 1, 1, 1, 1 } };  

solveMaze(maze);  

return 0;  

}  

3 0
3 years ago
Which of the following kinds of computing refers to an environment of servers that house and provide access to resources users a
AveGali [126]

Answer:

Cloud is the correct answer for the above question.

Explanation:

Missing Information:

    The option is missing in the question and the question suggests to choice. The option can be liberated, cloud, disperse, digital for this question in which cloud is the correct answer.

Detailed Explanation:

  • The cloud is used to store a large amount of data or information for varies users and anyone can store or modify or delete the information from anywhere.
  • This service takes money from the user and gives service to modify, edit or add the data from anywhere using the internet. When a user wants to take the service of the cloud then he needs to contact the cloud service provider, then pays money to them and then he can enjoy the service.
  • The above question wants to ask about the term which is a vast environment of databases that provide access to the users to store and mane the data and that is the cloud which is the correct answer for the above question.
7 0
3 years ago
What would typically be the best camera position to achieve the most viewer engagement in the photograph?
sdas [7]

Answer:

Horizontal because it can achieve the widest possible view which in turn can also let your photo be larger.

5 0
3 years ago
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