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Sergio039 [100]
3 years ago
14

Write a method getIntVal that will get the correct value input as integer numbers from the user. The input will be validated bas

ed on the first two numbers received as parameters. In other words, your program will keep asking for a new number until the number that the user inputs is within the range of the and . The method should show a message asking for the value within the range as:
Computers and Technology
1 answer:
Setler [38]3 years ago
4 0

Answer:

 import java.util.Scanner;

public class Main

{  

   //Create a Scanner object to be able to get input

   public static Scanner input = new Scanner(System.in);

   

public static void main(String[] args) {

    //Ask the user to enter the lowest and highest values

    System.out.print("Enter the lowest: ");

    int lowest = input.nextInt();

    System.out.print("Enter the highest: ");

    int highest = input.nextInt();

   

    //Call the method with parameters, lowest and highest

    getIntVal(lowest, highest);

}

//getIntVal method

public static void getIntVal(int lowest, int highest){

    //Create a while loop

    while(true){

        //Ask the user to enter the number in the range

        System.out.print("Enter a number between " + lowest + " and " + highest + ": ");

        int number = input.nextInt();

       

        //Check if the number is in the range. If it is, stop the loop

        if(number >= lowest && number <= highest)

            break;

        //If the number is in the range,

        //Check if it is smaller than the lowest. If it is, print a warning message telling the lowest value

       

        if(number < lowest)

            System.out.println("You must enter a number that is greater than or equal to " + lowest);

        //Check if it is greater than the highest. If it is, print a warning message telling the highest value

        else if(number > highest)

            System.out.println("You must enter a number that is smaller than or equal to " + highest);

    }

}

}

Explanation:

*The code is in Java.

You may see the explanation as comments in the code

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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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A. True or false

1.True

2.false

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4.true

5.true

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