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Anit [1.1K]
3 years ago
15

There are 10 computers and 5 students. How many ways are there for the students to sit at the computers if no computer has more

than one student and each student is seated at a computer?
Mathematics
1 answer:
mestny [16]3 years ago
6 0

Answer:

30,240 ways

Step-by-step explanation:

This question is bothered on permutation. Permutation has to do with arrangement.

If there are 10 computers and 5 students, the number of ways students will sit at the computers if no computer has more than one student can be expressed as;

10P5 = 10!/(10-5)!

10P5 = 10!/(5)!

10P5 = 10*9*8*7*6*5!/5!

10P5 = 10*9*8*7*6

10P5 = 30,240

Hence the number of ways is 30,240 ways

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13.7*1/10=1.37 So far she has hiked 1.37 miles. 
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Answer:

4.8

Step-by-step explanation:

1% of 40 is 0.4.

So, just multiply 0.4 by 12 to get 4.8

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Should i do my teast for 12th grade calculus
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Answer:

attempt it, if it's too hard, sleep

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Step-by-step explanation:

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2 years ago
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For a fair coin, suppose you toss the coin 100 times.
Natasha_Volkova [10]

Using the normal distribution, it is found that there is a 0.0005 = 0.05% probability of getting more than 66 heads.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with \mu = np, \sigma = \sqrt{np(1-p)}.

For the binomial distribution, the parameters are given as follows:

n = 100, p = 0.5.

Hence the mean and the standard deviation of the approximation are given as follows:

  • \mu = np = 100(0.5) = 50.
  • \sigma = \sqrt{np(1-p)} = \sqrt{100(0.5)(0.5)} = 5

Using continuity correction, the probability of getting more than 66 heads is P(X > 66 + 0.5) = P(X > 66.5), which is <u>one subtracted by the p-value of Z when X = 66.5</u>.

Z = \frac{X - \mu}{\sigma}

Z = \frac{66.5 - 50}{5}

Z = 3.3

Z = 3.3 has a p-value of 0.9995.

1 - 0.9995 = 0.0005.

0.0005 = 0.05%

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

6 0
1 year ago
Suppose the average price of gasoline for a city in the United States follows a continuous uniform distribution with a lower bou
Novay_Z [31]

Answer:

P(X

And we can use the cumulative distribution function given by:

F(x) = \frac{x-a}{b-a} , a \leq X \leq b

And for this case we can write the probability like this:

P(X

And then the final answer for this case would be \frac{2}{3}=0.667

Step-by-step explanation:

For this case we define our random variable X "price of gasoline for a city in the USA" and we know the distribution is given by:

X \sim Unif (a=3.5, b=3.8)

And for this case the density function is given by:

f(x) = \frac{x}{b-a}= \frac{x}{3.8-3.5}=, 3.5 \leq X \leq 3.8

And we want to calculate the following probability:

P(X

And we can use the cumulative distribution function given by:

F(x) = \frac{x-a}{b-a} , a \leq X \leq b

And for this case we can write the probability like this:

P(X

And then the final answer for this case would be \frac{2}{3}=0.667

5 0
3 years ago
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