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Komok [63]
1 year ago
15

Which of the following people is required by law to wear a seat belt? A. The operator of a truck weighing more than 26,000 lbs B

. A driver delivering newspapers on a home delivery route C. The driver of a truck manufactured in 1973 D. A passenger on a school bus
Computers and Technology
1 answer:
mars1129 [50]1 year ago
3 0
A, because you can't wear a seat belt if you are throwing newspaper out the car, school buses don't have seat belts and neither did trucks back in the 70's
You might be interested in
Write a function that implements another stack function, peek. Peek returns the value of the first element on the stack without
devlian [24]

Answer:

See explaination

Explanation:

StackExample.java

public class StackExample<T> {

private final static int DEFAULT_CAPACITY = 100;

private int top;

private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);

/**

* Returns a reference to the element at the top of this stack.

* The element is not removed from the stack.

* atreturn element on top of stack

* atthrows EmptyCollectionException if stack is empty

*/

public T peek() throws EmptyCollectionException

{

if (isEmpty())

throw new EmptyCollectionException("stack");

return stack[top-1];

}

/**

* Returns true if this stack is empty and false otherwise.

* atreturn true if this stack is empty

*/

public boolean isEmpty()

{

return top < 0;

}

}

//please replace "at" with the at symbol

Note:

peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.

3 0
2 years ago
(Displaying a Sentence with Its Words Reversed) Write an application that inputs a line of text, tokenizes the line with String
Serjik [45]

Answer:

I am writing a JAVA and Python program. Let me know if you want the program in some other programming language.

import java.util.Scanner; // class for taking input from user

public class Reverse{ //class to reverse a sentence

public static void main(String[] args) { //start of main() function body

Scanner input = new Scanner(System.in); //create a Scanner class object

   System.out.print("Enter a sentence: "); //prompts user to enter a sentence

   String sentence = input.nextLine(); // scans and reads input sentence

   String[] tokens = sentence.split(" "); // split the sentence into tokens using split() method and a space as delimiter

   for (int i = tokens.length - 1; i >= 0; i--) { //loop that iterates through the tokens of the sentence in reverse order

       System.out.println(tokens[i]); } } } // print the sentence tokens in reverse order

Explanation:

In JAVA program the user is asked to input a sentence. The sentence is then broken into tokens using split() method. split method returns an array of strings after splitting or breaking the sentence based on the delimiter. Here the delimiter used is space characters. So split() method splits the sentence into tokens based on the space as delimiter to separate the words in the sentence. Next the for loop is used to iterate through the tokens as following:

Suppose the input sentence is "How are you"

After split() method it becomes:

How

are

you

Next the loop has a variable i which initialized to the tokens.length-1. This loop will continue to execute till the value of i remains greater than or equals to 0.

for (int i = tokens.length - 1; i >= 0; i--)

The length of first token i.e you is 3 so the loop executes and prints the word you.

Then at next iteration the value of i is decremented by 1 and now it points at the token are and prints the word are

Next iteration the value of i is again decremented by i and it prints the word How.

This can be achieved in Python as:

def reverse(sentence):

   rev_tokens = ' '.join(reversed(sentence.split(' ')))

   return rev_tokens

   

line = input("enter a sentence: ")

print((reverse(line)))

The method reverse takes a sentence as parameter. Then rev_tokens = ' '.join(reversed(sentence.split(' ')))  statement first splits the sentence into array of words/tokens using space as a delimiter. Next reversed() method returns the reversed sentence.  Next the join() method join the reversed words of the sentence using a space separator ' '.join(). If you want to represent the reversed tokens each in a separate line then change the above statement as follows:

rev_tokens = '\n'.join(reversed(sentence.split(' ')))  

Here the join method joins the reversed words separating them with a new line.

Simply input a line of text and call this reverse() method by passing the input line to it.

4 0
1 year ago
Which phrase best describes a scenario in Excel 2016?
Airida [17]

Answer:

what are the phrases?

Explanation:

6 0
1 year ago
An employee sets up an automation that transfers files in a specific folder on their PC to a remote drive for archiving, provide
Yuri [45]

Answer:

Rule based automation

Explanation:

8 0
1 year ago
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

5 0
1 year ago
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