Answer:
Here is the C++ program:
#include<iostream> //to use input output functions
#include<iomanip> //to use setprecision
using namespace std; //to identify objects cin cout
int main() { //start of main function
int n1, n2, n3, n4, n5, n6, n7, min, max; // declare variables for 7 numbers, minimum value and maximum value
double sum= 0; //declare variable to hold sum of 7 numbers
double average; //declare variable to hold average of 7 numbers
cout<<"Enter first number: "; //prompts user to enter 1st number
cin>>n1; //reads first number from user
max = n1; //sets the first number to maximum
min=n1; //sets the first number to minimum
cout<<"Enter second number: "; //prompts user to enter 2nd number
cin>>n2; //reads second number from user
if(n2<min){ //if second number is less than min
min=n2; } //sets min to n2
if(n2>max){ //if n2 is greater than max
max = n2; } //sets max to n2
cout<<"Enter third number: "; //prompts user to enter 3rd number
cin>>n3; //reads third number from user
if(n3<min){ //checks if n3 is greater than min
min=n3; } //sets n3 to min
if(n3>max){ //checks if n3 is greater than max
max = n3; } //sets max to n3
cout<<"Enter fourth number: ";//prompts user to enter 4th number
cin>>n4; //reads fourth number from user
if(n4<min){ //if n4 is less than min
min=n4; } //sets min to n4
if(n4>max){ //if n4 is greater than max
max = n4; } //sets max to n4
cout<<"Enter fifth number: "; //prompts user to enter 5th number
cin>>n5; //reads fifth number from user
if(n5<min){ //if n5 is less than min
min=n5; } //sets min to n5
if(n5>max){ //if n5 is greater than max
max = n5; } //sets max to n5
cout<<"Enter sixth number: "; //prompts user to enter 6th number
cin>>n6; //reads sixth number from user
if(n6<min){ // if n6 is less than min
min=n6; } //sets min to n6
if(n6>max){ //if n6 is greater than max
max = n6; } //sets max to n6
cout<<"Enter seventh number: ";//prompts user to enter 7th number
cin>>n7; //reads seventh number from user
if(n7<min){ //if n7 is less than minimum number
min=n7; } //assigns n7 to min
if(n7>max){ //if n7 is greater than the maximum number
max = n7; } //assigns n7 to max
sum = n1+n2+n3+n4+n5+n6+n7; //adds 7 numbers
average = sum/7; //computes average of 7 numbers
cout<<"The average is: "<<fixed<<setprecision(3)<<average<<endl; //displays average value up to show 3 digits below decimal point using setprecision method of iomanip library
cout<<"The maximum number is: "<<max<<endl; //displays maximum number of 7 numbers
cout<<"The minimum number is: "<<min<<endl; //displays miimum number of 7 numbers
return 0; }
Explanation:
The program is well explained in the comments attached to each statement of program. For example if
n1 = 3
n2 = 9
n3 = 7
n4 = 6
n5 = 2
n6 = 5
n7 = 4
When n1 is read using cin then this number is set to max and min as:
min = 9
max = 3
Now when n2 is read, the first if condition checks if n2 is less than min and second if condition checks if n2 is greater than max. As n2 = 9 so it is not less than min so this if condition is false and n2 is greater than max i.e. 3 so this condition is true. So 9 is assigned to max.
min = 9
max = 9
Now when n3 is read the values of min and max become:
min = 7
max = 9
Now when n4 is read the values of min and max become:
min = 6
max = 9
Now when n5 is read the values of min and max become:
min = 2
max = 9
Now when n6 is read the values of min and max become:
min = 2
max = 9
Now when n7 is read the values of min and max become:
min = 2
max = 9
Now the statement sum = n1+n2+n3+n4+n5+n6+n7;
executes which becomes:
sum = 3 + 9 + 7 + 6 + 2 + 5 + 4
sum = 36.0
Next program control moves to statement:
average = sum/7;
this becomes
average = 36/7
5.142857
Since this is to be displayed up to 3 decimal places so average = 5.143
the complete output of the program is attached.
What are you asking for
Answer:
The solution is provided in the explanation section.
Detailed explanation is provided using comments within the code
Explanation:
import java.util.*;
public class Main {
//The Bubble sort method
public static void bb_Sort(int[] arr) {
int n = 10; //Length of array
int temp = 0; // create a temporal variable
for(int i=0; i < n; i++){
for(int j=1; j < (n-i); j++){
if(arr[j-1] > arr[j]){
// The bubble sort algorithm swaps elements
temp = arr[j-1];
arr[j-1] = arr[j];
arr[j] = temp;
}
}
}
}
public static void main(String[] args) {
//declaring the array of integers
int [] array = new int[10];
//Prompt user to add elements into the array
Scanner in = new Scanner(System.in);
//Use for loop to receive all 10 elements
for(int i = 0; i<array.length; i++){
System.out.println("Enter the next array Element");
array[i] = in.nextInt();
}
//Print the array elements before bubble sort
System.out.println("The Array before bubble sort");
System.out.println(Arrays.toString(array));
//Call bubble sort method
bb_Sort(array);
System.out.println("Array After Bubble Sort");
System.out.println(Arrays.toString(array));
}
}
Answer:
public class ArithmeticMethods
{
public static void main(String[] args) {
int number1 = 7;
int number2 = 28;
displayNumberPlus10(number1);
displayNumberPlus10(number2);
displayNumberPlus100(number1);
displayNumberPlus100(number2);
displayNumberPlus1000(number1);
displayNumberPlus1000(number2);
}
public static void displayNumberPlus10(int number){
System.out.println(number + 10);
}
public static void displayNumberPlus100(int number){
System.out.println(number + 100);
}
public static void displayNumberPlus1000(int number){
System.out.println(number + 1000);
}
}
Explanation:
Inside the main:
Initialize two integers, number1 and number2
Call the methods with each integer
Create a method called displayNumberPlus10() that displays the sum of the given number and 10
Create a method called displayNumberPlus100() that displays the sum of the given number and 100
Create a method called displayNumberPlus1000() that displays the sum of the given number and 1000
Answer:
Code is provided in the attachment form
Explanation:
Vector Multiplication code:
Vector_multiplication.c file contains C code.
See attachment no 1 attached below
Classification of Algorithm: For creating elements in vector c of size n, number of scalar multiplications is equal to n. The size of original two vectors scales directly with the number of operations. Hence, it is classified as O(n).
Answer:
web browser
Explanation:
the other three will not let her excess her page at all
Answer:
Abstraction refines concepts to their core values, stripping away ideas to the fundamentals of the abstract idea. It leaves the common details of an idea. Abstractions make it easier to understand code because it concentrates on core features/actions and not on the small details.
This is only to be used for studying purposes.
Hope it helps!