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3 years ago

What is a geam in the ggplot2 system?

Computers and Technology

1 answer:

riadik2000 [5.3K]3 years ago

3 0

Answer:

a) a plotting object like point, line, or other shape

Explanation:

A geom in the ggplot2 system is "a plotting object like point, line, or other shape."

A Geom is used in formulating different layouts, shapes, or forms of a ggplot2 such as bar charts, scatterplots, and line diagrams.

For example some different types of Geom can be represented as geom_bar(), geom_point(), geom_line() etc.

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In order to generate a random number, you must use Math.random( ). Group of answer choices True False

Tanzania [10]

Answer:

False

Explanation:

While Math.random() can be used to generate a random number, Java programming language also has a class called Random in the java.util package which can be imported into your code. This is a more efficient way of generating random numbers (ints or doubles) as you can instantiate several random number generators. The following line of code creates an an object of the class Random and sets the bound to 10.

Random rand = new Random (10);

This will generate random number from 0-9.

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4 years ago

Explain briely what this statement mean. "A byte is equivalent to a character"

dedylja [7]

To put it briefly, a byte is equivalent to a character in that it encodes a single character, being this in the form of a letter, number, or symbol.

A byte is the smallest unit of storage memory on any modern computer. This byte is commonly made up of eight bits, a combination of binary digits used to represent data. The hierarchy of computer memory is as follows:

1 byte
1 kilobyte
1 megabyte
1 gigabyte
1 terabyte

The statement "A byte is equivalent to a character" is quite literal in its meaning given that through the use of the bits that comprise it, a byte is used to represent and store the data for a single character of text, being that a letter, number or at times a symbol.

To learn more:

brainly.com/question/13188094?referrer=searchResults

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3 years ago

A selected graphic appears surrounded by a(n) ______, which has small squares and circles around its edges.

Elza [17]

Sizing handles :))))))))))

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3 years ago

Which of the following is used to allocate memory for the instance variables of an object of a class?1. the reserved word public

kodGreya [7K]

Answer:

The correct answer to the following question will be 2. the operator new.

Explanation:

New operator is used to allocating the memory to the instance object.The new object can be created by using a "new" keyword in java .

Syntax of using 'new' operator is :

class_name object_name=new class_name() // it allocated the memory to the class

For Example :

ABC obj = new ABC;

Now, this time obj points to the object of the ABC class.

obj = new ABC ();

call the construction of ABC class

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3 years ago

The running time of Algorithm A is (1/4) n2+ 1300, and the running time of another Algorithm B for solving the same problem is 1

Mnenie [13.5K]

Answer:

Answer is explained below

Explanation:

The running time is measured in terms of complexity classes generally expressed in an upper bound notation called the big-Oh ( "O" ) notation. We need to find the upper bound to the running time of both the algorithms and then we may compare the worst case complexities, it is also important to note that the complexity analysis holds true (and valid) for large input sizes, so, for inputs with smaller sizes, an algorithm with higher complexity class may outperform the one with lower complexity class i.e, efficiency of an algorithm may vary in cases where input sizes are smaller & more efficient algorithm might be outperformed by the lesser efficient algorithms in those cases.

That's the reason why we consider inputs of larger sizes when comparing the complexity classes of the respective algorithms under consideration.

Now coming to our question for algorithm A, we have,

let F(n) = 1/4x² + 1300

So, we can tell the upper bound to the function O(F(x)) = g(x) = x2

Also for algorithm B, we have,

let F(x) = 112x - 8

So, we can tell the upper bound to the function O(F(x)) = g(x) = x

Clearly, algorithmic complexity of algorithm A > algorithmic complexity of algorithm B

Hence we can say that for sufficiently large inputs , algorithm B will be a better choice.

Now to find the exact location of the graph in which algorithmic complexity for algorithm B becomes lesser than

algorithm A.

We need to find the intersection point of the given two equations by solving them:

We have the 2 equations as follows:

y = F(x) = 1/4x² + 1300 __(1)

y = F(X) = 112x - 8 __(2)

Let's put the value of from (2) in (1)

=> 112x - 8 = 1/4x² + 1300

=> 112x - 0.25x² = 1308

=> 0.25x² - 112x + 1308 = 0

Solving, we have

=> x = (112 ± 106) / 0.5

=> x = 436, 12

We can obtain the value for y by putting x in any of the equation:

At x=12 , y= 1336

At x = 436 , y = 48824

So we have two intersections at point (12,1336) & (436, 48824)

So before first intersection, the

Function F(x) = 112x - 8 takes lower value before x=12

& F(x) = 1/4x² + 1300 takes lower value between (12, 436)

& F(x) = 112x - 8 again takes lower value after (436,∞)

Hence,

We should choose Algorithm B for input sizes lesser than 12

& Algorithm A for input sizes between (12,436)

& Algorithm B for input sizes greater than (436,∞)

8 0

3 years ago