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DIA [1.3K]
3 years ago
8

Assume that we have seeded a program with 5 defects before testing. After the test, 20 defects were detected, of which, 2 are fr

om the seeded defects and 18 are real, non-seeded defects. We know there are 3 more seeded defects remaining in the software. Assuming a linear relationship, what is the estimated real, non-seeded defects that may still be remaining in the program
Computers and Technology
1 answer:
melamori03 [73]3 years ago
5 0

Answer:

see explaination

Explanation:

Given that we have;

Defects before testing = Defects planted = 5

Defects after testing = 20

Seeded defects found = 2

Real, non seeded defects = 18

Hence, Total number of defects = (defects planted / seeded defects found) * real, non seeded defects = (5/2)*18 = 45

Therefore, Estimated number of real defects still present = estimated total number of defects - number of real, non seeded defects found = 45-18 = 27

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Elis [28]

Answer:

There is no need to make an algorithm for this simple problem. Just add the two numbers by storing in two different variables as follows:

Let a,b be two numbers.

c=a+b;

print(c);

But, if you want to find the sum of more numbers, you can use any loop like for, while or do-while as follows:

Let a be the variable where the input numbers are stored.

while(f==1)

{

printf(“Enter number”);

scanf(“Take number into the variable a”);

sum=sum+a;

printf(“Do you want to enter more numbers? 1 for yes, 0 for no”);

scanf(“Take the input into the variable f”);

}

print(Sum)

Explanation:

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3 years ago
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Explanation:

8 0
3 years ago
Which browser do most web users use and why do you think that it is that way? Make sure you search online for statistics to conf
oksian1 [2.3K]

Answer:

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4 years ago
1. Declare two dimensional double array with 3 rows and 4 columns. a) Use scanf to read the elements for this array. b) Write a
Kobotan [32]

Answer:

Following are the program in the C Programming Language:

#include <stdio.h> //header file

float avgs(int a[3][4]){ //define function

 int s =0, c=0; //set integer type variable

 float avg=0;   //set float type variable

 //set for loops to find sum of an array

 for (int i = 0; i < 3; i++){

   for (int j = 0; j < 4; j++){

   s = s + a[i][j];

   c +=1;

   }

 }

 //find average of an array  

 avg = s / c;

 return avg;

}

int min(int a[3][4]){//define function

 int m = a[0][0];  //set integer type variable  

 //set for loop for traversing

 for (int i = 0; i < 3; i++)

 {  for (int j = 0; j < 4; j++){

     if(m > a[i][j])

       m = a[i][j];

   }

 }

 return m;

}

int main(){//define main function

 int a[3][4];//set integer type Array

 printf("Enter the elements of an array:\n");

 for (int i = 0; i < 3; i++)

  for (int j = 0; j < 4; j++)

     scanf("%d", &a[i][j]);

 printf("\n Average of an array: %f", avgs(a));

 printf("\n Smallest element of an array: %d", min(a));

 return 0;

}

Explanation:

Here, we define header file "stdio.h" for print or scan the data.

Then, we define float type function "avgs()" for find the average of an array inside it.

  • we set two integer type variable for sum and count
  • we set float type variable to store average of an array
  • we set the for loops for the sum of an array.
  • we divide sum from count to find average and return it.

Then, we define integer type function "min()" to find the minimum value of an array.

  • we set integer type variable to store array elements
  • we set the for loops for traversing the array
  • we set if statement to find minimum value of an array.
  • we return the minimum value of an array.

Then, we define integer type "main()" function to call both the functions.

  • we set integer type array variable and get input from the user in it with the help of for loops.
  • we call both the functions with message and return 0.

8 0
4 years ago
Write a program for a grocery store to calculate the total charge for customers. In the main: Your program should ask customer t
gavmur [86]

Answer:

Hello Joelwestwood! This is a good question to check your knowledge of subrountines and Arrays. Please find the implementation with the explanation below.

Explanation:

The solution can be implemented in a number of languages including C++, Java and Python to name a few. Though the implementation language is not specified in the question, i'll provide you the implementation code in Python and also in Java. Java implementation is a little complex because we need to define an array size to be able to add items to it, whereas Python allows us more flexibility by allowing dynamic size of Array.

JAVA IMPLEMENTATION

import java.util.Scanner;

import java.util.Arrays;

class TotalCharge {

 private static int[] price_array = new int[1];

 public void FillPriceArray(int price) {

   if (price_array.length == 1) {

     price_array[0] = price;

   } else {

     price_array = new int[price_array.length + 1];

     price_array[price_array.length + 1] = price;

   }

 }

 public int[] get_price_array() {

   return price_array;

 }

 public static void main(String args[]) {

   System.out.println("Please enter the total number of items to purchase: ");

   Scanner scan = new Scanner(System.in);

   int price = scan.nextInt();

   if (price <= 20) {

     System.out.println("Items being purchased are less than 20");

   } else {

     System.out.println("Items being purchased are more than 20");

   }

   TotalCharge itemsTotal = new TotalCharge();

   itemsTotal.FillPriceArray(price);

   System.out.println(Arrays.toString(itemsTotal.get_price_array()));

 }

}

PYTHON IMPLEMENTATION

calculate_total_charge.py

def FillPriceArray(price):

 price_array.append(price)

price_array = []

price = raw_input("Please enter the total number of items to purchase: ")

try:

 price = int(price)

 if price <= 20:

   print("Items being purchased are less than 20")

 else:

   print("Items being purchased are more than 20")

 FillPriceArray(price)

 print(price_array)

except ValueError:

 print("Invalid input! Exiting..")

8 0
3 years ago
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