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7nadin3 [17]
1 year ago
13

A wire 22 cm long has a mass of 374 g. What is the mass of 13 cm of this wire?

Mathematics
1 answer:
taurus [48]1 year ago
7 0

Answer:

<u>2</u><u>2</u><u>1</u><u> </u><u>g</u>

The up pinned pic is of inverse variation typed answer.. If u want word problem type answer here are the steps (EVEN IF THE STEPS ARE DIFFERENT ANSWERS REMAIN SAME)

Step-by-step explanation:

Mass of wire from 22cm = 374g

Mass of wire from 1 cm = 374÷22 = 17g

Mass of wire from 13 cm = 13×17 = <u>2</u><u>2</u><u>1</u><u> </u><u>g</u>

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Answer: 9

Step-by-step explanation:

In order to find the people out of the 150, we first multiply 9/25 and 150 together. This will give us the people who DO own a dog. This brings us to 54.

Now, if we multiply 54 by 1/6 we get the answer of 9, because 1/6 of the dog owners own a cat as well. We have now reached the desired amount.

6 0
1 year ago
According to a Pew Research survey, about 27% of American adults are pessimistic about the future of marriage and the family. Th
IgorLugansk [536]

Answer:

P(X≤5)=0.5357

Step-by-step explanation:

Using the binomial model, the probability that x adults from the sample, are pessimistic about the future is calculated as:

P(x)=\frac{n!}{x!(n-x)!} *p^{x}*(1-p)^{n-x}

Where n is the size of the sample and p is the probability that an adult is pessimistic about the future of marriage and family. So, replacing n by 20 and p by 0.27, we get:

P(x)=\frac{20!}{x!(20-x)!}*0.27^{x}*(1-0.27)^{20-x}

Now, 25% of 20 people is equal to 5 people, so the probability that, in a sample of 20 American adults, 25% or fewer of the people are pessimistic about the future of marriage and family is equal to calculated the probability that in the sample of 20 adults, 5 people of fewer are pessimistic about the future of marriage and family.

Then, that probability is calculated as:

P(X≤5)= P(1) + P(2) + P(3) + P(4) + P(5)

Where:

P(0)=\frac{20!}{0!(20-0)!}*0.27^{0}*(1-0.27)^{20-0}=0.0018

P(1)=\frac{20!}{1!(20-1)!}*0.27^{1}*(1-0.27)^{20-1}=0.0137

P(2)=\frac{20!}{2!(20-2)!}*0.27^{2}*(1-0.27)^{20-2}=0.0480\\P(3)=\frac{20!}{3!(20-3)!}*0.27^{3}*(1-0.27)^{20-3}=0.1065\\P(4)=\frac{20!}{4!(20-4)!}*0.27^{4}*(1-0.27)^{20-4}=0.1675\\P(5)=\frac{20!}{5!(20-5)!}*0.27^{5}*(1-0.27)^{20-5}=0.1982

Finally, P(X≤5) is equal to:

P(X≤5) = 0.0018+0.0137 + 0.0480 + 0.1065 + 0.1675 + 0.1982

P(X≤5) = 0.5357

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