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2 years ago

8

Find the area of the shaded region.

1 answer:

yaroslaw [1]2 years ago

4 0

It’s ether 138.6 or 101.4. 123 rounded to the nearest tenth is 120 so you ether add or subtract

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Here you go this is the full question please help

hammer [34]

I don’t see anything is there supposed to be a picture or?

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2 years ago

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What is The value of -8/14 · 20/64

Genrish500 [490]

The answer is -5/28

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2 years ago

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PLEASE HELP!! WILL GIVE BRAINLIEST FOR BEST ANSWER!! Please solve #17 and #18. If you can only answer one it's ok.

Paha777 [63]

Well for number 17 you have to multiply by two which 4 * 2 is 8 and 12 * 2= 24. for number 18 your answer would be isometry. hope that helped

5 0

2 years ago

Find y when x = 15 if y = 6 when x= 30

Mandarinka [93]

y=3 when x=15

this is because we can set up a ratio of when x/15 and set it equal to 6/30 and divide them, solve for x.

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3 years ago

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, de

Lelechka [254]

Answer:

(a) $P(X=10) = 0.2070$

(b) $P(X\geq 10) = 0.3798$

Step-by-step explanation:

Note that in this problem we have an initial population N = 50, of which 30 fulfill a certain characteristic "m" (belong to the second section). Then, from the population N, a sample of size n = 15 is selected and it is desired to know how many comply with the desired characteristic (second section).

So

Let X be the number of projects in the second section, then X is a discrete random variable that can be modeled by a hypergeometric distribution.

(a)

Therefore, to answer question (a) we use the following equation presented in the attached image:

Where:

$N = 50\\m = 30\\n = 15\\X = 10$

Then:

$P (X = 10) = \frac{\frac{30!}{10!(30-10)!}\frac{20!}{5!(20-5)!}}{\frac{50!}{15!(50-15)!}}\\\\\\P(X=10) = 0.2070$

(b)

For part (b) we have:

$P(X\geq10) = 1-P(X$

$P(X\geq 10) = 1-0.6202 \\\\P(X\geq 10) = 0.3798$

4 0

3 years ago