Why do you need to put your phone on airplane mode.

Pro and Cons of Artificial Intelligence in Art You must have 3 statements in each

Anni [7]

Answer:

The answer is below

Explanation:

Aritifiaicla intelligence in art is an artwork created by the application of artificial intelligence software.

Some of the pros of artificial intelligence in the art are:

1. It creates a new and improved interface, specifically in graphic design such as virtual reality and 3D printing

2. It creates and mixes the artistic ideas properly, such as mixing of different instruments in music to creates a new sound

3. It establishes graphical and visual display with no blemishes o,r error when applied accordingly, such as AUTOCAD

The cons of artificial intelligence in art are:

1. Artificial Intelligence lacks emotional sense. Hence it is challenging to display artistic elements that portray genuine emotions

2. It lacks creativity. Unlike humans, artificial intelligence is not as creative as humans when it comes to words or sentence constructions in an artistic sense.

3. It doesn't apply experience to its productions. Arts can be improved with more understanding of what is happening in the society or environment; artificial intelligence cannot translate its experience into arts formation.

6 0

3 years ago

Write the definition of a function max that has three int parameters and returns the largest.

g100num [7]

Solution:

The definition of a function max that has three int parameters and returns the largest is given bellow:

def max(x,y,z):

if (x>z and x>y):

return (x)

elif (y>x and y>z):

return y

else:

return z

Thus this is required right answer.

4 0

3 years ago

Read 2 more answers

write a pseudo code and flow chart that take a number as input and prints multiplication table up to 10

Troyanec [42]

PSEUDOCODE:

1. DECLARE number: INTEGER

2. DECLARE multiple: INTEGER

3. INPUT number

4. FOR counter FROM 1 TO 10 DO

5. multiple <-- number * counter

6. PRINT number, " * ", counter, " = ", multiple

7. ENDFOR

1. declaring a variable "number" as an Integer

2. declaring a variable "multiple" as an Integer

3. The user inputs the value of number

4. FOR loop where variable "counter" increments by 1 after every iteration

5. sets the value for variable "multiple" as the value of number * counter

6. prints out for example "3 * 1 = 3" and will continue till counter reaches 10

7. Ends the for loop

FLOWCHART below

hope it helped

8 0

3 years ago

Assume that word is a variable of type String that has been assigned a value. Assume furthermore that this value always contains

Valentin [98]

Answer:

String word = "George slew the dragon";

int pos = word.indexOf("dr");

String drWord = word.substring(pos, pos+4);

System.out.println(drWord);

Explanation:

Assuming dr is always there, we don't have to check the validity of 'pos'. Normally, you would!

5 0

4 years ago

Read 2 more answers

A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0

3 years ago