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Where can i check on an acer chromebook to see what version of java is installed.

lana66690 [7]

Based on the current situation, to check on an Acer Chromebook to see what version of java is installed, you must "input specific commands on the terminal window and press Enter key."

<h3>Java Installation on Chromebook.</h3>

Currently, Góogle or Chromebook do not allow Java installation on its operating system. This is due to the security threats that are associated with Java.

However, there are ways to bypass this situation and eventually install Java on your Chromebook if you so desire.

Nevertheless, if eventually, you installed Java successfully on your Chromebook, to check the version, you must "input specific commands on the terminal window and press Enter key."

Learn more about Java Program here: brainly.com/question/19485769

5 0

2 years ago

What does the following code print?double[] myList = {1, 5, 5, 5, 5, 1};double max = myList[0];int indexOfMax = 0;for (int i = 1

Liula [17]

Answer:

This code will print: 4

Explanation:

Following is the step-by-step explanation for the given code:

Given is the array of data type double named myList, it has entries, 1, 5, 5, 5,5, 1:

double[] myList = {1, 5, 5, 5, 5, 1};

Now the first element of the array (1) with index 0 will be stored in the variable max (data type double).

double max = myList[0];

A variable indexOfMax having datatype int will be initiated as 0.

int indexOfMax = 0;

Now for loop will be used to find the maximum number of the array. The variable i will be put as index for each element to compare with first element. If the checked element is greater than or equal to the integer in max, it will be replaced. So at the end the variable max will have value 5 that will be at index i = 4.

for (int i = 1; i < myList.length; i++)

{ if (myList[i] >= max)

{ max = myList[i];

Now the variable i that is the index for max value will be stored in the variable indexOfMax (indexOfMax = 4).

indexOfMax = i; }}

At end the value stored in variable indexOfMax will be printed, so 4 will be printed as output.

System.out.println(indexOfMax);

i hope it will help you!

7 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 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 from a set of 13 cards.

Explanation:

The crucial point 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 combination 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 combinations is $\\ \frac{n!}{(n-k)!k!}$ , where n is the total of elements available and k is the size of a selection of k elements from which we can choose from the total n.

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 Multiplication Principle 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 multiplication of all them.

5 0

4 years ago

List the different generation of computers with its main component

Dennis_Churaev [7]

Answer:

1940 – 1956: First Generation

1956 – 1963: Second Generation

1964 – 1971: Third Generation

1972 – 2010: Fourth Generation

2010- : (Present )Fifth Generation

Explanation:

First Generation Computers (1940-1956):In this Generation the main component of computers were Vacuum Tubes.
Second Generation Computers (1956-1963):In this generation the main component of computers were Transistors.
Third Generation Computers (1964-1971):In this generation the main component of computers were Integrated Circuits.
Fourth Generation Computers (1972-2010):In this generation the main component of computers were Microprocessor
Fifth Generation (2010-Present):In this generation the main component of computers is Artificial Intelligence

4 0

3 years ago

Assume you have 100 values that are all different, and use equal width discretization with 10 bins.

zepelin [54]

Answer:

a) 10

b) 1

C) 10

D) 1

E) 20

F) 10

Explanation:

a) The largest number of records that could appear in one bin

= 10

B) The smallest number of records that could appear in one bin

= 1

C) The largest number of records that cab appear in one bin

= 10

d) smallest number

= 1

e) With frequency = 20. the largest number of records that could appear in one bin with equal width discretization (10 bins)

= 20

f ) with equal height discretization

= 10

6 0

3 years ago