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Ainat [17]
2 years ago
6

Write a program that reads a list of integers, and outputs whether the list contains all even numbers, odd numbers, or neither.

The input begins with an integer indicating the number of integers that follow.If the input is:5246810Then the output is:all evenIf the input is:513579
Computers and Technology
2 answers:
astraxan [27]2 years ago
8 0

Answer:

uuuuu

Explanation:

nadya68 [22]2 years ago
6 0

Answer:

57677748957394209690369-34064666

Explanation:

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Get user input as a boolean and store it in the variable tallEnough. Also, get user input as a boolean and store it in the varia
frozen [14]

Answer:

I will code in JAVA.

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

 

boolean tallEnough;

boolean oldEnough;

Scanner input = new Scanner(System.in);

tallEnough = input.nextBoolean();<em> //wait the input for tallEnough</em>

oldEnough = input.nextBoolean(); <em>//wait the input for OldEnough</em>

   if(tallEnough && oldEnough){

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   }

}

}

Explanation:

First, to accept user inputs you have to import the class Scanner. Then declare both variables before allowing the user to set input values for both boolean variables.

In the if-else statement checks if both variables are true, then prints true. Another case prints always false.

8 0
3 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec.  b) 5.12 msec

Explanation:

If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:

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The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:

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⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

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