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Rasek [7]
3 years ago
13

What is the number of ways to arrange 5 objects from a set of 8 different objects?

Mathematics
1 answer:
jekas [21]3 years ago
4 0

Answer:

6,720 ways

Step-by-step explanation:

Since in the problem arrangemnt is being asked this is a problem of permutation.

No . of ways of arranging r things out of n things is given by

P(n,r) =   n!/(n-r)!

In the problem given we have to arrange 5 objects from set of 8 objects.

Here n = 8 and p = 5

it can be done in in

P(8,5) =   8!/(8-5)! ways

8!/(8-5)!  = 8!/3! = 8*7*6*5*4*3!/3! = 8*7*6*5*4 = 6,720

Thus,  number of ways to  arrange 5 objects from a set of 8

different objects is P(8,5) =   8!/(8-5)! =  6,720 .

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Classify square root 36 A. Whole Number B. Integer C. Rational Number D. Irrational Number
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There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In
Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the <em>sampling distribution of the sampling proportions</em>.

The mean and standard deviation of this sampling distribution is:

\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

P(0.45

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

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For the second game, we win if we get a proportion over 80%.

The probability of winning is:

P(p>0.8)=P(z>z^*)

The z value is calculated as before:

z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0

As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).

If our chances of winnings depend on P(z>z*), they become lower as z* increases.

Then, we can conclude that our chances of winning decrease with the increase of the number of trials.

We prefer the option of 30 trials for this game.

8 0
3 years ago
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