Need help fixing my code!!
Keep getting an error code that says my member is inaccessible

An example of a leading question is: "Did Dorian ever get into fights in second grade?" Therefore, option C is correct.

Leading questions are questions that are framed in a way that suggests or encourages a particular answer or direction. They are designed to influence the respondent's perception or show their response toward a desired outcome. Leading questions can unintentionally or intentionally bias the answers given by the person being questioned.

Leading questions may include specific words or phrases that guide the respondent toward a particular answer.

Learn more about leading questions, here:

https://brainly.com/question/31105087

#SPJ2

Answer:

keyword involved

Explanation:

simple select the desired text and click SHIFT + F3

a. Proposition 2 states that Insertion-Sort runs in O(n + X) time, where X is the number of inversions.

b. The expected number of inversions, E[X],  E[X] = 1/4n(n-1).

c. In the average case, Insertion-Sort has a time complexity of approximately O(1/4n²).

(a) Proposition 2 states that Insertion-Sort runs in O(n + X) time, where X is the number of inversions. To understand why this is true, let's refer to the pseudocode for Insertion-Sort:

InsertionSort(A):

  for i from 1 to length[A] do

     key = A[i]

     j = i - 1

     while j >= 0 and A[j] > key do

        A[j + 1] = A[j]

        j = j - 1

     A[j + 1] = key

b. The expected number of inversions, E[X], can be calculated as follows:

E[X] = Σ(i,j) E[I(i, j)]

= Σ(i,j) Pr((i, j) is an inversion)

= Σ(i,j) 1/2

= (n(n-1)/2) * 1/2

= n(n-1)/4

Hence, E[X] = 1/4n(n-1).

(c) Using Proposition 2 and the result from part (b), we can determine the average case time complexity of Insertion-Sort. The average case time complexity is given by O(n + E[X]).

Substituting the value of E[X] from part (b):

Average case time complexity = O(n + 1/4n(n-1))

Simplifying further:

Average case time complexity = O(n + 1/4n^2 - 1/4n)

Since 1/4n² dominates the other term, we can approximate the average case time complexity as:

Average case time complexity ≈ O(1/4n²)

Therefore, in the average case, Insertion-Sort has a time complexity of approximately O(1/4n²).

Learn more about proposition on

https://brainly.com/question/30389551

Here's an example program in C++ that fulfills the requirements:

_______________________________________________________

#include <iostream>

#include <fstream>

#include <vector>

using namespace std;

int main() {

   const int ARRAY_SIZE = 1000; // Set a large size for the array

   double array[ARRAY_SIZE];

   double sum = 0.0;

   int count = 0;

   // Read the first five elements from the keyboard

   cout << "Enter the first five elements of the array:\n";

   for (int i = 0; i < 5; i++) {

       cout << "Element " << i + 1 << ": ";

       cin >> array[i];

       sum += array[i];

       count++;

   }

   // Read the remaining elements from a file

   ifstream inputFile("input.txt"); // Replace "input.txt" with your file name

   double num;

   while (inputFile >> num && count < ARRAY_SIZE) {

       array[count] = num;

       sum += array[count];

       count++;

   }

   inputFile.close();

   // Calculate and print the average

   double average = sum / count;

   cout << "Average of array elements: " << average << endl;

   return 0;

}

________________________________________________________

In this C++ program, a large array of doubles is declared with a size of 1000. The first five elements are read from the keyboard using a for loop, and the sum and count variables keep track of the cumulative sum and the number of elements entered.

The remaining elements are read from a file named "input.txt" (you should replace it with the actual file name) using an ifstream object. The program continues reading elements from the file as long as there are more numbers and the count is less than the array size.

Finally, the average is calculated by dividing the sum by the count, and it is printed to the console. Remember to replace "input.txt" with the correct file name and double-check the average with a calculator to ensure accuracy.

~~~Harsha~~~

Answer:

Load and save instructions. The specific instructions may vary depending on the computer`s architecture, but the general process is to load the desired value into a register and store it in a flip-flop. Below is an example of a hypothetical assembly procedure.  

Load Instruction: Load the value 1 into a register.

"""

LOAD R1, 1

"""

Store Instruction: Store the value from the register into the flip-flop.

"""

STORE R1, FlipFlop

"""
weird question but this might help