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BartSMP [9]
2 years ago
12

A variable like userNum can store a value like an integer. Extend the given program to print userNum values as indicated.(1) Out

put the user's input.Enter integer: 4You entered: 4(2) Extend to output the input squared and cubed. Hint: Compute squared as userNum * userNum. (Submit for 2 points, so 4 points total).Enter integer: 4You entered: 44 squared is 16 And 4 cubed is 64!! (3) Extend to get a second user input into userNum2. Output sum and product. (Submit for 1 point, so 5 points total).Enter integer: 4You entered: 44 squared is 16 And 4 cubed is 64!!Enter another integer: 54 + 5 is 94 * 5 is 20LABACTIVITY1.16.1: Basic output with variables (Java)0 / 5OutputWithVars.javaLoad default template...import java.util.Scanner;public class OutputWithVars {public static void main(String[] args) {Scanner scnr = new Scanner(System.in);int userNum = 0;System.out.println("Enter integer: ");userNum = scnr.nextInt(); return;}}import java.util.Scanner;public class OutputWithVars {public static void main(String[] args) {Scanner scnr = new Scanner(System.in);int userNum = 0;System.out.println("Enter integer: ");userNum = scnr.nextInt();return;}}Develop modeSubmit modeRun your program as often as you'd like, before submitting for grading. Below, type any needed input values in the first box, then click Run program and observe the program's output in the second box.
Computers and Technology
1 answer:
dimulka [17.4K]2 years ago
8 0

Answer:

The program in Java is as follows:

import java.util.*;

public class Main{

public static void main(String[] args) {

 int userNum;

 Scanner input = new Scanner(System.in);

 System.out.print("Enter integer: ");

 userNum = input.nextInt();

 System.out.println("You entered "+userNum);

 System.out.println(userNum+" squared is "+(userNum*userNum));

 System.out.println(userNum+" cubed is "+(userNum*userNum*userNum));

 int userNum2;

 System.out.print("Enter another integer: ");

 userNum2 = input.nextInt();

 int sum= userNum + userNum2; int product = userNum2 * userNum;

 System.out.println(userNum+" + "+userNum2+" = "+sum);

 System.out.println(userNum+" * "+userNum2+" = "+product); } }

Explanation:

This declares userNum

 int userNum;

 Scanner input = new Scanner(System.in);

This prompts the user for input

 System.out.print("Enter integer: ");

This gets user input from the user

 userNum = input.nextInt();

Number (1) is implemented here

 System.out.println("You entered "+userNum);

Number (2) is implemented here

 System.out.println(userNum+" squared is "+(userNum*userNum));

 System.out.println(userNum+" cubed is "+(userNum*userNum*userNum));

This declares another variable userNum2

 int userNum2;

This prompts the user for another input

 System.out.print("Enter another integer: ");

This gets user input from the user

 userNum2 = input.nextInt();

This calculates the sum and the product

 int sum= userNum + userNum2; int product = userNum2 * userNum;

Number (3) is implemented here

 System.out.println(userNum+" + "+userNum2+" = "+sum);

 System.out.println(userNum+" * "+userNum2+" = "+product);

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Dima020 [189]

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

8 0
3 years ago
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