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Mama L [17]
3 years ago
13

(Java) Meadowdale Dairy Farm sells organic brown eggs to local customers. It charges $3.25 for a dozen eggs, or 45 cents for ind

ividual eggs that are not part of a dozen. Write a program that prompts a user for the number of eggs in the order and then display the amount owed with a full explanation using the following wording: You ordered 27 eggs. That’s 2 dozen at $3.25 per dozen and 3 loose eggs at 45 cents each for a total of $7.85.
import java.util.Scanner;
class Eggs {
public static void main(String[] args) {
// Write your code here

}
}​
Computers and Technology
1 answer:
zepelin [54]3 years ago
8 0

import java.util.Scanner;

public class Eggs {

   

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.println("How many eggs?");

       int eggs = scan.nextInt();

       int dozens = eggs / 12, ind = eggs - (dozens*12);

       

       System.out.println("You ordered "+eggs+" eggs. That's "+dozens+" at $3.25 per dozen and "+ind+" loose eggs at 45 cents each for a total of $"+((dozens*3.25)+(ind*0.45))+".");

   }

   

}

I hope this helps!

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Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

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Explanation:

a.) A packet can make 1,2 or 3 hops

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Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

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if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

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   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

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