Answer:
Java code is given below
Explanation:
import java.util.Random;
class Die{
private int sides;
public Die(){
sides = 6;
}
public Die(int s){
sides = s;
}
public int Roll(){
Random r = new Random();
return r.nextInt(6)+1;
}
}
class DieRoll{
public static void main(String[] args) {
Die die = new Die();
int arr[] = new int[6];
for(int i=0; i<6; i++)
arr[i] = 0;
for(int i=0; i<100; i++){
int r = die.Roll();
arr[r-1]++;
}
for(int i=0; i<6; i++)
System.out.println((i+1)+" was rolled "+arr[i]+" times.");
}
}
Answer:
#include <stdio.h>
void interchangeCase(char phrase[],char c){
for(int i=0;phrase[i]!='\0';i++){
if(phrase[i]==c){
if(phrase[i]>='A' && phrase[i]<='Z')
phrase[i]+=32;
else
phrase[i]-=32;
}
}
}
int main(){
char c1[]="Eevee";
interchangeCase(c1,'e');
printf("%s\n",c1);
char c2[]="Eevee";
interchangeCase(c2,'E');
printf("%s\n",c2);
}
Explanation:
- Create a function called interchangeCase that takes the phrase and c as parameters.
- Run a for loop that runs until the end of phrase and check whether the selected character is found or not using an if statement.
- If the character is upper-case alphabet, change it to lower-case alphabet and otherwise do the vice versa.
- Inside the main function, test the program and display the results.
Make the zig zag part more spaced out
Answer:
1. Input() is the correct answer
2.Most programming languages have a data type called a string, which is used for data values that are made up of ordered sequences of characters, such as "hello world". A string can contain any sequence of characters, visible or invisible, and characters may be repeated. ... A string can be a constant or variable .
3.The int() function converts the specified value into an integer number.
4.CandyCost = int(input("How much is the candy?"))
5.int()
ANSWERS
Answer:
The answer is option A.
Explanation:
Negative numbers can be found by binary search, this makes option B incorrect.
Unsorted and randomized lists are also not things that support a binary search, options C and D are incorrect.
Binary search uses a technique where the middle element of the list is located and used to determine whether the search should be done within the lower indexed part of the list or the higher. So for a list to be binary search-able, it should be sorted and not randomized.
The answer is A.
I hope this helps.
