If you're unsure if something

This is a free point thing!! I am giving 100 points!! Anyone wanna talk!!!!

julia-pushkina [17]

Answer:

:)

Explanation:

8 0

3 years ago

Read 2 more answers

Group of answer choices When declaring a variable, you also specify the type of its values. Variables cannot be assigned and dec

olga nikolaevna [1]

Complete Question:

Which of the following statement is true?

Group of answer choices.

A. When declaring a variable, you also specify the type of its values.

B. Variables cannot be assigned and declared in the same statement.

C. Variable names can be no more than 8 characters long.

D. Variable names must contain at least one dollar sign.

Answer:

A. When declaring a variable, you also specify the type of its values.

Explanation:

In Computer programming, a variable can be defined as a placeholder or container for holding a piece of information that can be modified or edited.

Basically, variable stores information which is passed from the location of the method call directly to the method that is called by the program.

For example, they can serve as a model for a function; when used as an input, such as for passing a value to a function and when used as an output, such as for retrieving a value from the same function. Therefore, when you create variables in a function, you can can set the values for their parameters.

For instance, to pass a class to a family of classes use the code;

\\parameter Name as Type (Keywords) = value;

\\procedure XorSwap (var a,b :integer) = "myvalue";

Hence, the true and correct statement is that when declaring a variable, you also specify the type of its values such as integers, string, etc.

6 0

2 years ago

Assume that printStars is a function that takes one argument and returns no value. It prints a line of N stars (followed by a ne

trasher [3.6K]

Answer:

printStars(35);

Explanation:

public class Question {

public static void main(String args[]) {

printStars(35);

}

public static void printStars(int numberOfStars){

for(int i = 1; i <= numberOfStars; i++){

System.out.print("*");

}

System.out.print("\n");

}

7 0

3 years ago

In most programming languages, the compiler carries a preprocessing step to determine if certain statements will compile. For in

MakcuM [25]

Answer:

See explaination for program code

Explanation:

package com.company;

import java.io.IOException;

import java.nio.file.Files;

import java.nio.file.Paths;

import java.util.Stack;

/*

* A program which reads input from given text filename and prints

* whether it passed the pre-processing stage or not

* Sample input

* int A = (a + b);static void display(int x)

* {

* //A sample input file

* }

* */

public class MyPreprocessor {

/*

* Reads input from filename and returns a string

* containing whole text

* */

public String readInput(String filename) {

String content = "";

try {

content = new String(Files.readAllBytes(Paths.get(filename)));

} catch (IOException e) {

e.printStackTrace();

}

return content;

}

/**

* Filters input since we are only concerned with

* a subset of input = '{}[]()/*'

*/

public String filterText(String text) {

String output = "";

for(int i = 0;i<text.length();++i) {

switch (text.charAt(i)) {

case '[':

case ']':

case '{':

case '}':

case '(':

case ')':

case '*':

case '/':

output += text.charAt(i);

}

return output;

}

/*

* Uses stack to determine if input is valid or not

* */

public boolean parse(String input) {

Stack<Character> St = new Stack<Character>();

// '$' is special symbol to represent bottom of stack

St.push('$');

for(int i = 0;i<input.length();++i) {

Character symbol = input.charAt(i);

if(St.peek().equals(getOpenSymbol(symbol)))

St.pop();

else St.push(symbol);

}

return St.peek() == '$';

}

private static Character getOpenSymbol(Character symbol) {

switch (symbol) {

case '}': return '{';

case ']': return '[';

case ')': return '(';

case '*': return '*';

case '/': return '/';

default: return ' ';

}

public static void main(String[] args) {

MyPreprocessor preprocessor = new MyPreprocessor();

String inputText = preprocessor.readInput("src/input.txt");

String filteredInput = preprocessor.filterText(inputText);

if(preprocessor.parse(filteredInput))

System.out.println("preprocessing passed.");

else System.out.println("preprocessing failed!");

}

6 0

3 years ago

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

8 0

2 years ago