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KatRina [158]
2 years ago
13

Describe some common types of charts.​

Computers and Technology
2 answers:
UkoKoshka [18]2 years ago
6 0

Answer:

The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things.

Explanation:

Line chart- A graph with points connected by lines to show how something changes in value: • as time goes by, • or as something else changes.

Example: how you are improving at a quiz each day.

Bar chart- A graph drawn using rectangular bars to show how large each value is. The bars can be horizontal or vertical. It is a really good way to show relative sizes. We can use bar graphs to show the relative sizes of many things, such as what type of car people have, how many customers a shop has on different days and so on.

Histogram- a graphical display of data using bars of different heights. It is similar to a Bar Chart, but a histogram groups numbers into ranges . The height of each bar shows how many fall into each range.

Pie chart- a special chart that uses "pie slices" to show relative sizes of data. It is a really good way to show relative sizes.

Cartesian chart- This is a type of chart that uses coordinates to pinpoint where something is on a map or graph. Using Cartesian Coordinates we mark a point on a graph by how far along and how far up it is.

Mekhanik [1.2K]2 years ago
4 0
Bar chart.
Pie chart.
Line chart.
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Consider the recursive method whose definition appear below. Why? public static String mysteryString (String s){ if(s.length()==
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Answer:

retupmoc

Explanation:

1.) Anwser will be retupmoc

because

public static String mysteryString(String s){

if(s.length() == 1){

return s;

}

else{

return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));

}

}

In this program input is "computer" . So the function mysteryString(String s) it does

return s.substring(s.length() -1) + mysteryString(s.substring(0, s.length()-1));

so when it enters the first time ??s.substring(s.length() -1) and it will be give you 'r' then it calls the function recursively by reducing the string length by one . So next time it calls the mysteryString function with string "compute" and next time it calls return s.substring(s.length()-1)? + mysteryString(s.substring(0,s.length-1)) so this time it gives "e" and calls the function again recursively . It keeps on doing till it matched the base case.

so it returns "retupmoc".

6 0
3 years ago
Decribe the components of an Microsoft<br>Windows desktop<br>​
damaskus [11]

Explanation:

the main component of Windows when you start your computer are the desktop, my computer, recycle bin , start button, Taskbar and shortcuts to application after opening an application you will also have a Windows with an active application which will appear in your desktop and taskbar.

3 0
3 years ago
A sql-6-5.sql file has been opened for you. Write each of the following tasks as a SQL statement in a new line (remember that yo
Ghella [55]

Answer:

The query is as follows:

select sum(stock) as total_stock from products

Explanation:

Required

Return total stock using the alias total_stock from the product table.

The explanation of the query is as follows:

select ----> This implies that data is to be selected from the table

sum(stock) ----> This adds up entries in stock column

as total_stock ---> This represents the alias used for sum(stock)column where

from products  ----> The table being queried

Take for instance, the content of the table is:

SN  Product Stock

1      Apple     5

2     Orange   3

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The query will return the following table:

total_stock

16

5 0
2 years ago
Create a linked list for library patrons. Put at least 8 patrons in the list. (You should hard code this data.) Ask for a person
Evgesh-ka [11]

Answer:

Explanation:

#include <bits/stdc++.h>

using namespace std;

/* Link list node */

class Node

{

public:

string Name;

long Library_card_number;

string Street;

string City;

long Zip;

Node* next;

};

/* Given a reference (pointer to pointer) to the head

of a list and an int, push a new node on the front

of the list. */

void push(Node** head_ref,string name,long lcn,string street,string city,long zip)

{

/* allocate node */

Node* new_node = new Node();

/* put in the key */

new_node->Name = name;

new_node->Library_card_number = lcn;

new_node->Street = street;

new_node->City = city;

new_node->Zip = zip;

/* link the old list off the new node */

new_node->next = (*head_ref);

/* move the head to point to the new node */

(*head_ref) = new_node;

}

/* Checks whether the value x is present in linked list */

Node* search(Node* head, string x)

{

Node* current = head; // Initialize current

while (current != NULL)

{

if (current->Name == x)

return current;

current = current->next;

}

return NULL;

}

/* Driver program to test count function*/

int main()

{

/* Start with the empty list */

Node* head = NULL;

int x = 21;

/* Use push() to construct below list

14->21->11->30->10 */

push(&head, "Joe",1,"hippy Street","Bay area",2009);

push(&head, "Marie",2,"lippy street","san jose",2010);

push(&head, "Harry",3,"gippy Street","Bay area",2009);

push(&head, "Ashish",4,"dippy street","san jose",2010);

push(&head, "Zuck",5,"sippy Street","Bay area",2009);

push(&head, "Gates",6,"kippy street","san jose",2010);

push(&head, "Page",7,"pippy Street","Bay area",2009);

push(&head, "Brin",8,"dippy street","san jose",2010);

string name;

cout<<"Enter name of the person to search: ";

cin>>name;

Node* z = search(head,name);

if(z!=NULL){

cout<<"\nPerson found in the list"<<endl;

cout<<"Name: "<<z->Name<<endl;

cout<<"Library Card Number: "<<z->Library_card_number<<endl;

cout<<"Street: "<<z->Street<<endl;

cout<<"City: "<<z->City<<endl;

cout<<"Zip: "<<z->Zip<<endl;

}else{

cout<<"No person found with the name "<<name<<endl;

}

return 0;

}

7 0
2 years ago
A software engineer is developing a new operating system. The operating system has to make a decision when it sees an applicatio
anyanavicka [17]

Answer:

No.

Be patient. Wait a bit longer to see if the application succeeds, and if it does not, force quit and debug the operating system to figure out what's wrong.

7 0
2 years ago
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