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alexandr402 [8]
3 years ago
9

Which formula should Gemima use to show the

Computers and Technology
2 answers:
statuscvo [17]3 years ago
7 0

Answer:

it is =SUM(ABOVE)......

Explanation:

Shalnov [3]3 years ago
3 0
The answer is
Sum(above)
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What kind of device is a printer? output or input
ANEK [815]
A printer is output. :)
8 0
3 years ago
A related database stores data in the form of______
avanturin [10]

Answer:

Lists.

Explanation:

A relational database stores data in the form of lists.

6 0
2 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
3 years ago
Consider the following algorithms. Each algorithm operates on a list containing n elements, where n is a very large integer.
DochEvi [55]

We have that the appropriate statement will be

  • An algorithm that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the algorithms run in reasonable time

III only.

Option B

From the question we are told

Consider the following <u>algorithms</u>. Each <u>algorithm</u> operates on a list containing n <em>elements</em>, where n is a very large <u>integer</u>.

I. An algorithm that accesses each <u>element</u> in the list twice.

II. An <em>algorithm </em>that accesses each <u>element </u>in the list n times.

III. An <u>algorithm</u> that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the <em>algorithms </em>run in <em>reasonable </em>time?

<h3>Algorithm  </h3>

Generally In order to get <em>admission </em>to every thing in the list twice, the algorithm has to traverse the listing twice,

which leads to 2*n entry to operations.

And if the every factor is accessed n times, the listing will be traversed n time, which leads to n^2 get right of entry to operations.

If n is a very giant <em>integer</em>, each 2*n and n^2 are plenty larger.

So, there will be <em>solely </em>ten entry to operations and this algorithm will have a sensible jogging time.

Therefore

An algorithm that accesses only the first 10 elements in the list, regardless of the size of the list. Which of the algorithms run in reasonable time

III only.

Option B

For more information on  algorithm  visit

brainly.com/question/950632

4 0
2 years ago
What is it called when you define a variable for the first time
Nimfa-mama [501]

Answer:

That is called declaring a variable

Explanation:

6 0
2 years ago
Read 2 more answers
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