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4 years ago

6

If you like to spend time outdoors working with plants and animals, you have a(n) _____. a. bodily/kinesthetic learning style b.

intrapersonal learning style c. visual/spatial learning style d. naturalistic learning style

Computers and Technology

2 answers:

torisob [31]4 years ago

7 0

Naturalistic learning style

Varvara68 [4.7K]4 years ago

7 0

The answer is D.) naturalistic learning style

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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.

5 0

4 years ago

What is the "key" to a Caesar Cipher that someone needs to know (or discover) to decrypt the message? a) A secret word only know

Tpy6a [65]

Answer:

The number of characters to shift each letter in the alphabet.

Explanation:

Caeser Cipher is the technique of encryption of data, to make it secure by adding characters between alphabets. These are the special characters that make the message secure while transmitting.

According to the standards, For Decryption, we remove these special characters between alphabets to make message understandable.

<em>So, we can say that,to de-crypt the message, the number of characters to shift each letter in the alphabet.</em>

3 0

3 years ago

5. The value of a variable

Volgvan

Answer:

It's true .

The value of a variable

cannot be changed during

program execution.

6 0

4 years ago

Carol typed a memo to send to everyone in her department. To create this memo, she used a _____.

aleksley [76]

she used a word processor

8 0

4 years ago

Read 2 more answers

Active and passive security measures are employed to identify detect classify and analyze possible threats inside of which zone?

allsm [11]

Answer:

The answer is "Assessment Zone"

Explanation:

An Assessment zone provides different types of safety zones that are active and passive measures to identify, detect, classify and analyze potential threats that are employed within the evaluation zones.

It is routing in wireless sensor networks based on hierarchical routing.

The assessment zone provides illustrates the validity of the model and the algorithm.

4 0

4 years ago