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rusak2 [61]
3 years ago
14

What is the GCF of 55 and 24

Mathematics
1 answer:
KIM [24]3 years ago
6 0

Answer:

1 is the GCF of 55 and 24

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A public health organization reports that 40%of baby boys 6-8 months old in the United
Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

  • 0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.
  • 0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.
  • Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

P(X > 4) = 1 - P(X \leq 4)

In which:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Then:

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

P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061

P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403

P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150

P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502

Hence:

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106

P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016

P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

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1 year ago
PLEASE look at photo and help!
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6n- 15 = 6(5)? – 15<br> =
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what

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2 years ago
Find the third number if their average is 6 and two of them are 6 and 7​
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Answer:

5

Step-by-step explanation:

If their average is 6 and there are three numbers in total, the total sum of the three numbers would be:

6×3 = 18

You know that two of them is 6 and 7, so you can find the last number easily.

18-6-7 = 5  

8 0
3 years ago
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