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allochka39001 [22]
3 years ago
13

Write a program that allows the user to convert a temperature given in degrees from either Celsius to Fahrenheit or Fahrenheit t

o Celsius. Use the following formulas: Degrees_C = 5(Degrees_F− 32)/9 and Degrees_F = (9(Degrees_C)/5) + 32) Prompt the user to enter a temperature and either a C or c for Celsius or an F or f for Fahrenheit. Convert the temperature to Fahrenheit if Celsius is entered, or to Celsius if Fahrenheit is entered. Display the result in a readable format. If anything other than C, c, F, or f is entered, print an error message and stop.
Computers and Technology
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

import java.util.Scanner;

public class num6 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter the temperature");

       double temp = in.nextDouble();

       System.out.println("Enter C to convert to celsius or F for fahrenheit");

       String con = in.next();

       if (con.equalsIgnoreCase("C")) {

           double Degrees_C = 5 * (temp - 32) / 9;

           System.out.println("The temperature of " + temp + "F in Celcius is" +

                   " " + Degrees_C);

       } else if (con.equalsIgnoreCase("F")) {

           //Degrees_F = (9(Degrees_C)/5) + 32)

           double Degrees_F = (9 * (temp) / 5) + 32;

           System.out.println("The temperature of " + temp + " celcius in Farenheit is" +

                   " " + Degrees_F);

       } else {

           System.out.println("Invalid value for conversion");

       }

   }

}

Explanation:

Use Scanner Class to prompt and receive user input

Use string method equalsIgnoreCase in an if statement to test the conversion to degrees celcius or farenheit

evaluate and output the values based on the equation given in the question

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Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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