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Snezhnost [94]
3 years ago
14

State two features of a word processor​

Computers and Technology
1 answer:
Diano4ka-milaya [45]3 years ago
3 0

Answer:

Some of the functions of word processing software include:

Creating, editing, saving and printing documents.

Copying, pasting, moving and deleting text within a document.

Formatting text, such as font type, bolding, underlining or italicizing.

Creating and editing tables.

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>

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The preferred means of creating multithreaded Java applications is by implementing the ________ interface. An object of a class
Kaylis [27]

Answer:

Runnable.

Explanation:

Java is an all round programming language which is typically object-oriented and class-based. It was designed by James Gosling, developed by Sun microsystems and released officially on the 23rd of May, 1995. Java programming language is designed to have only a few implementation dependencies as possible because it was intended to be written once, and run on any platform.

Java makes concurrency to be available to software developers through the application programming interface (API) and the language. Also, it supports multiple threads of execution, by making each thread have its respective program counter and method-call stack, which then allow concurrent executions with other threads.

The preferred means of creating multithreaded Java applications is by implementing the runnable interface. An object of a class that implements this interface represents a task to perform. The code is public void run ().

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3 years ago
C programming Write a function named CalculateSphereVolume that takes one integer parameter intDiameter. The function should cal
FinnZ [79.3K]

Answer:

not familiar with C++. but basically save the constant 4/3 in a variable and use user input of cin << i  believe to ask for a radius. Then take that radius/input saved in a variable and cube it, then return the value.

Explanation:

i hope this works.

3 0
3 years ago
A person gets 13 cards of a deck. Let us call for simplicity the types of cards by 1,2,3,4. In How many ways can we choose 13 ca
natima [27]

Answer:

There are 5,598,527,220 ways to choose <em>5</em> cards of type 1, <em>4 </em>cards<em> </em>of type 2, <em>2</em> cards of type 3 and <em>2</em> cards of type 4 from a set of 13 cards.

Explanation:

The <em>crucial point</em> of this problem is to understand the possible ways of choosing any type of card from the 13-card deck.

This is a problem of <em>combination</em> since the order of choosing them does not matter here, that is, the important fact is the number of cards of type 1, 2, 3 or 4 we can get, no matter the order that they appear after choosing them.

So, the question for each type of card that we need to answer here is, how many ways are there of choosing 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 are of type 4 from the deck of 13 cards?

The mathematical formula for <em>combinations</em> is \\ \frac{n!}{(n-k)!k!}, where <em>n</em> is the total of elements available and <em>k </em>is the size of a selection of <em>k</em> elements  from which we can choose from the total <em>n</em>.

Then,

Choosing 5 cards of type 1 from a 13-card deck:

\frac{n!}{(n-k)!k!} = \frac{13!}{(13-5)!5!} = \frac{13*12*11*10*9*8!}{8!*5!} = \frac{13*12*11*10*9}{5*4*3*2*1} = 1,287, since \\ \frac{8!}{8!} = 1.

Choosing 4 cards of type 2 from a 13-card deck:

\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-4)!4!} = \frac{13*12*11*10*9!}{9!4!} = \frac{13*12*11*10}{4!}= 715, since \\ \frac{9!}{9!} = 1.

Choosing 2 cards of type 3 from a 13-card deck:

\\ \frac{n!}{(n-k)!k!} =\frac{13!}{(13-2)!2!} = \frac{13*12*11!}{11!2!} = \frac{13*12}{2!} = 78, since \\ \frac{11!}{11!}=1.

Choosing 2 cards of type 4 from a 13-card deck:

It is the same answer of the previous result, since

\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-2)!2!} = 78.

We still need to make use of the <em>Multiplication Principle</em> to get the final result, that is, the ways of having 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 is the multiplication of each case already obtained.

So, the answer about how many ways can we choose 13 cards so that there are 5 of type 1, there are 4 of type 2, there are 2 of type 3 and there are 2 of type 4 is:

1287 * 715 * 78 * 78 = 5,598,527,220 ways of doing that (or almost 6 thousand million ways).

In other words, there are 1287 ways of choosing 5 cards of type 1 from a set of 13 cards, 715 ways of choosing 4 cards of type 2 from a set of 13 cards and 78 ways of choosing 2 cards of type 3 and 2 cards of type 4, respectively, but having all these events at once is the <em>multiplication</em> of all them.

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4 years ago
What is SoC? how is it different from CPU?
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Answer:

SoC usually contains a GPU

Explanation:

Where as a CPU cannot function without dozens of other chips

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