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2 years ago

Is(are) input value(s) to use with a program to test the accuracy of the output.

Computers and Technology

2 answers:

Law Incorporation [45]2 years ago

5 0

Answer:

test cases

Explanation:

Readme [11.4K]2 years ago

5 0

I think it’s the third one hope that helped

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I don't know the answer and I tried different ones and they not what the book is looking for.

Serhud [2]

Answer:

The complete program is:

import java.util.Scanner;

class Main

{

public static void main (String [] args) {

int numObjects;

Scanner scnr = new Scanner(System.in);

numObjects = scnr.nextInt(); // Program will be tested with values: 15, 40.

System.out.println(numObjects);

}

Explanation:

The program is as mentioned above, And we can check with inputs 15, and 40. And it was found that correct answer is obtained on running the program.

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Answer:

Explanation:

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2 years ago

Group of programs are software

Cloud [144]

Answer:

Explanation:

a single program or application is a software

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2 years ago

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I have a bag of trail mix.half of the bag contains peanuts.1/4of the bag is chocolates candies, and 1/4of the bag is dried fruit

nika2105 [10]

A.) 1/2 Half the bags are peanuts so you have 1/2 a chance of selecting s peanut.

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2 years ago

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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

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 $n = 15\ and\ r = 4$

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

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

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

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

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

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

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)

From the given parameters; 3 out of 15 is detective;

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;

Probability that at least one is defective + Probability that at none is defective = 1

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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2 years ago