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Vedmedyk [2.9K]
3 years ago
11

Suppose that a t-shirt comes in five sizes: S, M, L, XL, XXL. Each size comes in four colors. How many possible t-shirts are the

re?
Computers and Technology
1 answer:
guapka [62]3 years ago
3 0

Answer:

120

Explanation:

This question is basically asking how many ways can we arrange 5 sizes of shirts if they come in 4 colors.

5 permutation of 4

To find permutation, we use the formula

P(n, r) = n! / (n - r)!

Where

n is the size of the shirts available, 5

r is the color of each shirt available, 4

P(5, 4) = 5! / (5 - 4)!

P(5, 4) = (5 * 4 * 3 * 2 * 1) / 1!

P(5, 4) = 120 / 1

P(5, 4) = 120

Therefore, the total number of shirts available is 120

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The replacer parameter of the stringify method accepts a/an ______________ or an array.
Elina [12.6K]

Answer:

The answer is function

Explanation:

As a function, it accepts two parameters: the value and the key being stringified.  It can be used to change the value of the entries or eliminate them. As an array, it specifies by name which entries and names of the properties in the object to include in the resulting JSON string.

7 0
3 years ago
If you anticipate running a particular query often, you can improve overall performance by saving the query in a special file ca
mestny [16]

If you anticipate running a particular query often, you can improve overall performance by saving the query in a special file called a(n)  stored procedure.

<h3>What is meant by stored procedure?</h3>

An application that uses a relational database management system can access a subroutine known as a stored procedure. These processes are kept in the data dictionary of the database. Access control and data validation are two applications for stored processes.

A stored procedure is a collection of SQL statements that have been given a name and are kept together in a relational database management system (RDBMS) so they may be used and shared by various programs.

If you expect to run a certain query frequently, putting it in a special file known as a(n) stored procedure can enhance overall speed.

To learn more about stored procedure refer to:

brainly.com/question/13692678

#SPJ4

3 0
2 years ago
Given the following narrative for Bambino’s Pizzeria, list the actors that should be used in the use-case diagram.
Andreyy89

Answer:

See explaination

Explanation:

Please kindly check attachment for the diagram in its best detailed manner.

8 0
4 years ago
Create a class called Hokeemon that can be used as a template to create magical creatures called Hokeeemons. Hokeemons can be of
Viefleur [7K]

Answer:

See Explaination

Explanation:

/*****************************************Hokeemon.java********************************/

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class Hokeemon {

private String name;

private String type;

private int age;

public Hokeemon(String name, String type, int age) {

super();

this.name = name;

this.type = type;

this.age = age;

}

public Hokeemon() {

this.name = "";

this.type = "";

this.age = 0;

}

public String getName() {

return name;

}

public void setName(String name) {

this.name = name;

}

public String getType() {

return type;

}

public void setType(String type) {

this.type = type;

}

public int getAge() {

return age;

}

public void setAge(int age) {

this.age = age;

}

public String liveIn() {

if (this.type.equalsIgnoreCase("dwarf")) {

return "Mountain";

} else if (this.type.equalsIgnoreCase("elf")) {

return "Dale";

} else if (this.type.equalsIgnoreCase("fairy")) {

return "Forest";

} else {

return "Shire";

}

}

public boolean areFriends(Hokeemon other) {

if (this.type.equalsIgnoreCase(other.type)) {

return true;

} else if ((this.type.equalsIgnoreCase("dwarf") && other.type.equalsIgnoreCase("elf"))

|| (this.type.equalsIgnoreCase("elf") && other.type.equalsIgnoreCase("dwarf"))) {

return true;

} else if ((this.type.equalsIgnoreCase("hobbit") && other.type.equalsIgnoreCase("fairy"))

|| (this.type.equalsIgnoreCase("fairy") && other.type.equalsIgnoreCase("hobbit"))) {

return true;

} else {

return false;

}

}

public static Hokeemon[] getData(String file) {

Hokeemon[] hokeemons = new Hokeemon[8];

int i = 0;

try {

Scanner scan = new Scanner(new File(file));

while (scan.hasNextLine() && i < hokeemons.length) {

String line = scan.nextLine();

String[] data = line.split("\\s+");

String name = data[0];

String type = data[1];

int age = Integer.parseInt(data[2]);

hokeemons[i] = new Hokeemon(name, type, age);

i++;

}

scan.close();

} catch (FileNotFoundException e) {

System.err.println(e);

}

return hokeemons;

}

public static String getBio(Hokeemon[] hokeemons) {

String s = "";

for (Hokeemon hokeemon : hokeemons) {

s += "I am " + hokeemon.getName() + ": Type " + hokeemon.getType() + ": Age=" + hokeemon.getAge()

+ ", I live in " + hokeemon.liveIn() + "\n";

s += "My friends are: ";

for (Hokeemon hokeemon2 : hokeemons) {

if (hokeemon.areFriends(hokeemon2) && !hokeemon.equals(hokeemon2)) {

s += (hokeemon2.getName() + " ");

}

}

s += "\n";

}

return s;

}

atOverride

public String toString() {

return "Name " + name + ": Type " + type + ": Age=" + age + "\n";

}

}

/*******************************HokeemonMain.java**************************/

import java.util.Arrays;

public class HokeemonMain {

public static void main(String[] args) {

Hokeemon[] hokeemons = Hokeemon.getData("data.txt");

System.out.println(Arrays.toString(hokeemons));

System.out.println(Hokeemon.getBio(hokeemons));

}

}

/**********************************data.txt********************/

Noddy Dwarf 4

Legolas Elf 77

Minerva Fairy 33

Samwise Hobbit 24

Merry Hobbit 29

Warhammer Dwarf 87

Ernedyll Elf 19

Frodo Hobbit 18

3 0
4 years ago
A contiguous subsequence of a list S is a subsequence made up of consecutive elements of S. For example, if S is 5,15,-30,10,-5,
Sindrei [870]

Answer:

We made use of the dynamic programming method to solve a linear-time algorithm which is given below in the explanation section.

Explanation:

Solution

(1) By making use of the  dynamic programming, we can solve the problem in linear time.

Now,

We consider a linear number of sub-problems, each of which can be solved using previously solved sub-problems in constant time, this giving a running time of O(n)

Let G[t] represent the sum of a maximum sum contiguous sub-sequence ending exactly at index t

Thus, given that:

G[t+1] = max{G[t] + A[t+1] ,A[t+1] } (for all   1<=t<= n-1)

Then,

G[0] = A[0].

Using the above recurrence relation, we can compute the sum of the optimal sub sequence for array A, which would just be the maximum over G[i] for 0 <= i<= n-1.

However, we are required to output the starting and ending indices of an optimal sub-sequence, we would use another array V where V[i] would store the starting index for a maximum sum contiguous sub sequence ending at index i.

Now the algorithm would be:

Create arrays G and V each of size n.

G[0] = A[0];

V[0] = 0;

max = G[0];

max_start = 0, max_end = 0;

For i going from 1 to n-1:

// We know that G[i] = max { G[i-1] + A[i], A[i] .

If ( G[i-1] > 0)

G[i] = G[i-1] + A[i];

V[i] = V[i-1];

Else

G[i] = A[i];

V[i] = i;

If ( G[i] > max)

max_start = V[i];

max_end = i;

max = G[i];

EndFor.

Output max_start and max_end.

The above algorithm takes O(n) time .

4 0
3 years ago
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