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8_murik_8 [283]
4 years ago
9

8) The higher the drag coefficient, the_____ the car will go a)Siower b)Faster

Computers and Technology
2 answers:
Murrr4er [49]4 years ago
6 0

the answer is * slower *

yanalaym [24]4 years ago
5 0
The correct answer is a)Slower
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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
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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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  • Learning is a process of connecting specialized nodes or information sources.
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  • Capacity to know more is more critical than what is currently known
  • Nurturing and maintaining connections is needed to facilitate continual learning.
  • Ability to see connections between fields, ideas, and concepts is a core skill.
  • Currency (accurate, up-to-date knowledge) is the intent of all connectivist learning activities.
  • Decision-making is itself a learning process. Choosing what to learn and the meaning of incoming information is seen through the lens of a shifting reality. While there is a right answer now, it may be wrong tomorrow due to alterations in the information climate affecting the decision
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