Answer:
See explaination
Explanation:
void showSeatingChart(string seatingChart[20][40], const int ROWS, const int COLS){
for(int i = 0;i<ROWS;i++){
for(int j = 0;j<COLS;j++){
cout<<seatingChart[i][j]<<" ";
}
cout<<endl;
}
}
Answer:
Yes Preto distribution takes over when we apply 80/20 rule.
Explanation:
The Pareto distribution is derived from pareto principle which is based on 80/20 rule. This is main idea behind the Pareto distribution that percent of wealth is owned by 20% of population. This 80/20 rule is gaining significance in business world. There it becomes easy to understand the input and output ratio. The input ratio helps to analyse the potential output using Pareto distribution.
Here is the JAVA program to find smallest distance between 2 neighboring numbers in an array.
import java.lang.Math; // for importing Math class functions
import java.util.Scanner; // for importing Scanner class
public class CalSmallestDistance // class to calculate smallest distance
{public static void main(String[] args) { // to enter java program
Scanner s = new Scanner(System.in); //creating scanner object
int size; // size of the array
System.out.print("Enter size of the array:"); //prompts to enter array size
size = s.nextInt(); // reads input
int arr[] = new int[size]; // array named arr[]
//line below prompts to enter elements in the array System.out.println("Enter numbers in the array:");
for(int j = 0; j < size; j++) //loops through the array
{arr[j] = s.nextInt(); } //reads the input elements
int smallest_distance = Math.abs(arr[0]-arr[1]);
int position= 0; //index of the array
for(int i=1; i<arr.length-1; i++){
int distance= Math.abs(arr[i]-arr[i+1]);
if(distance< smallest_distance){
smallest_distance= distance;
position = i; } }
System.out.println("Smallest distance is :"+smallest_distance);
System.out.println("The numbers are :"+arr[position]+ " and " +arr[position+1]); } }
I have stated the minor explanation of some basic lines of code as comments in the code given above.
Now i will give the detailed explanation about the working of the main part of the code.
Lets start from here
int smallest_distance = Math.abs(arr[0]-arr[1]);
In this statement the array element at 0 index (1st position) and the array element at 1 index (2nd position) of the array are subtracted.
Then i used the math.abs() method here which gives absolute value
Lets say the distance between 3 and 5 is -2 as 3-5 equals to -2. But the math.abs() method will return 2 instead of -2.
Now the subtraction of two array elements and absolute value (if subtraction result is negative) will be assigned to variable smallest_distance.
for(int i=1; i<arr.length-1; i++)
This is for loop. Variable i is positioned to the 1 index of the array (arr) (it is pointing to the second element of the array). It will move through the array until the end of the array is reached i.e. the loop will continue till value of i remains less than the length of the array.
Now at the start of the loop body the following statement is encountered
int distance= Math.abs(arr[i]-arr[i+1]);
This subtracts the array element at i th position and array element at i th +1 position (means one position ahead of array element at i th position). In simple words two neighboring numbers in an array are being subtracted and Math.abs() method is used to give absolute value. The result of this operation is assigned to distance variable.
if(distance< smallest_distance)
This if statement checks if the value in distance variable is smallest than the value of smallest_distance variable which was previously calculated before calculating the value for distance variable.
If this condition is true then the following statements are executed:
smallest_distance= distance;
if distance value is less than value in smallest_distance, then the value of distance is assigned to smallest_distance.
this means the smallest_distance will keep on storing the smallest distance between two neighboring numbers.
Next the value of variable i that is pointing to the 1st index of the array is now assigned to the position variable.
position = i;
It will keep assigning the value of i to position variable so at the end of the program we can get the positions of the two neighboring numbers that have the smallest distance between them.
Then the value of i is incremented and moves one place ahead in the array.
Then the 2nd iteration takes place and again checks if i pointer variable has reached the end of the array. If not the loop body will continue to execute in which the distance between the two neighboring numbers is calculated and shortest distance is stored in smallest_distance.
When i reaches the end of the array the loop will break and the smallest distance between two neighboring numbers in the array have been stored in the smallest_distance variable.
Finally the statement System.out.println("Smallest distance is :"+smallest_distance); displays the shortest distance and the statement System.out.println("The numbers are :"+arr[position]+ " and " +arr[position+1]); displays the array index positions at which the two neighboring numbers have the smallest distance.
Answer:
No, we do not recommend contracting without providing for full and open competition. There is sufficient evidence that the necessary BSVD is only available from the original source, NanoTech, and that use of any other contractor would create unacceptable delays in fulfilling the need.
B: the responsible person or organization and the website URL
C: the responsible person or organization, date accessed, and URL
D: the responsible person or organization and the date accessed
Answer:
C: the responsible person or organization, date accessed, and URL
Explanation:
Answer:
see explaination
Explanation:
I am considering some relationship for this ER Diagram as follows :
(i) One Car can have only one type.
(ii) One Car needs more than one repairings.
(iii) A customer can have more than one car and a car must belong to only one customer.
5 dequeue operations
6 enqueue operations
10 dequeue operations
8 enqueue operations
2 dequeue operations
3 enqueue operations
Last = 10
Answer:
10
Explanation:
An enqueue operation is a function that adds an element(value) to a queue array. A dequeue operations removes an element from a queue array. Queue arrays follow a first-in-first-out approach, so elements that are first stored in the queue are removed/accessed first(enqueue operations add elements at the rear of the queue array).
The following operations leave 10 elements in the queue of array size 12 after its done:
10 enqueue operations= adds 10 elements
5 dequeue operations= removes 5 elements( 5 elements left in queue)
6 enqueue operations= adds 6 elements(11 elements in queue)
10 dequeue operations= removes 10 elements(1 element left in queue)
8 enqueue operations= adds 8 elements(9 elements in queue)
2 dequeue operations= removes 2 elements(7 elements left in queue)
3 enqueue operations= adds 3 elements(10 elements in queue)
Therefore there are 10 elements in the queue after enqueue and dequeue operations.