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Svet_ta [14]
2 years ago
11

What is the sum of 9c+(8+3c)?

Mathematics
1 answer:
mixas84 [53]2 years ago
6 0
<span>9c+(8+3c)      
9c+8+3c     (collect like terms)

9c+3c+8
12c+8

12c+8 is the answer.
im not sure if your teacher also wants the answer to be in simplest form so im going to do that below as well:

12c+8      (4*3=12 and 4*2=8. you can take out a 4 from those 2 numbers)

4(3c+2) (this is the simplest form of </span>9c+(8+3c)<span>)


</span>
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Describe the features of the Free-market system
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At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is
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Answer:

34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

81% of students own a wireless device.

This means that p = 0.81

If 5 students are selected at random, what is the probability that all 5 have a wireless device?

This is P(X = 5) when n = 5. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{5,5}.(0.81)^{5}.(0.19)^{0} = 0.3487

34.87% probability that all 5 have a wireless device

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3 years ago
HELPPPPPPPP
Natasha2012 [34]
-$12.99
That should be it, hit me up for any additional questions
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3 years ago
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