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3 years ago

Explain how a set of numbers can have a greatest common factor greater than 1

Mathematics

1 answer:

g100num [7]3 years ago

5 0

The set of number can all be related to each other. They may all have more factors that are greater than 1.

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julie bought 32 kiwi fruit for $16. How many kiwi can lisa buy if she has $5?what is per one kiwi? 2 different examples

Gemiola [76]

The kiwi are 50 cents each. Therefore, Lisa can buy ten kiwi. What did you mean by the examples?

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3 years ago

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Lerok [7]

2 min and 7 sec plus 12= 2 min and 19 sec. Add 8 to the number 19, equals 26. It took Juan 2 min and 26 sec to finish.

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3 years ago

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posledela

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3 years ago

QUICK PLEASE ANSWER

Mila [183]

Answer:

63/80 and 7/4 * 9/20

Step-by-step explanation:

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2 years ago

g An urn contains 150 white balls and 50 black balls. Four balls are drawn at random one at a time. Determine the probability th

xxTIMURxx [149]

Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.

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.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

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

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

Sampling with replacement:

I consider a success choosing a black ball, so $p = \frac{50}{150+50} = \frac{50}{200} = 0.25$

We want 2 black balls and 2 white, 2 + 2 = 4, so $n = 4$ , and we want P(X = 2).

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

$P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109$

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so $N = 200$

Sample of 4, so $n = 4$

50 are black, so $k = 50$

We want P(X = 2).

$P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116$

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

3 0

2 years ago