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Phantasy [73]
3 years ago
5

Reduce then calculate 7/8 x 4/15 x 3/7 = ? Can you help?

Mathematics
1 answer:
Zepler [3.9K]3 years ago
8 0

since this is a multiplication, all the numerators are just factors of the product numerator and all denominators are just factors of the product denominator, so we can simply reorder them some, without changing the product.


\bf \cfrac{7}{8}\times \cfrac{4}{15}\times \cfrac{3}{7}\implies \cfrac{7}{7}\times \cfrac{4}{8}\times \cfrac{3}{15}\implies \cfrac{1}{1}\times \cfrac{1}{2}\times \cfrac{1}{5}\implies \cfrac{1\cdot 1\cdot 1}{1\cdot 2\cdot 5}\implies \cfrac{1}{10}

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3 years ago
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Let A = {June, Janet, Jill, Justin, Jeffrey, Jelly}, B = {Janet, Jelly, Justin}, and C = {Irina, Irena, Arena, Arina, Jelly}. Fi
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{June, Janet, Jill, Justin, Jeffrey, Jelly,Irina, Irena, Arena, Arina, }

Step-by-step explanation:

A ∪ C

This means union so we join the sets together

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A U C =  {June, Janet, Jill, Justin, Jeffrey, Jelly,Irina, Irena, Arena, Arina, Jelly}

We get rid of repeats

A U C =  {June, Janet, Jill, Justin, Jeffrey, Jelly,Irina, Irena, Arena, Arina, }

5 0
4 years ago
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8 0
3 years ago
Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.
lesya [120]

Answer:

0.1994 is the required probability.      

Step-by-step explanation:

We are given the following information in the question:

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Standard Deviation, σ = 5.3 pounds

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z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds

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3 years ago
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