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umka21 [38]
4 years ago
11

A dart hits the circular dartboard shown below at a random point. find the probability that the dart lands in the shaded square

region. the radius of the dartboard is 11in, and each side of the shaded region is 4in.
Mathematics
1 answer:
zaharov [31]4 years ago
7 0
The first thing you need to do is:you have a square whose area is 11^2 = 121 square inches.then we are going to multiply the radius to the PI which is 3.14 the area of the circle with radius 4 is equal to 3.14 * 4^2 = 3.14 * 16 = 50.24you need to divide the calculated radius times pi (3.14) you can get the answer and you need to round your answer to the nearest hundredththe probability that the dart will hit within the circle is equal to 50.24 / 121 = 0.4152066116 which is rounded to .04152 or 41.52%this assumes the dart will always hit the square at least.
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Answer:

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Step-by-step explanation:

The number of mice vary inversely with cats. This is written as:

Mice \propto \dfrac{1}{Cat} \\Mice = \dfrac{k}{Cat}

When there are 3r-19 cats, there are 2r+1 mice

2r+1 = \dfrac{k}{3r-19}\\$Cross multiply$\\k=(3r-19)(2r+1)

When there are 6r-27 mice, there are r-5 cats.

6r-27 = \dfrac{k}{r-5}\\$Cross multiply$\\k=(6r-27)(r-5)

Taking the values of k above, we have:

k=(3r-19)(2r+1) =(6r-27)(r-5)\\(3r-19)(2r+1) =(6r-27)(r-5)\\6r^2+3r-38r-19=6r^2-30r-27r+135\\$Collect like terms\\6r^2-6r^2+3r+30r-38r+27r-19-135=0\\22r-154=0\\22r=154\\$Divide both sides by 22\\r=7

Therefore:

k=(3r-19)(2r+1) \\=(3*7-19)(2*7+1)\\=(21-19)(14+1)\\=2*15\\k=30

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mezya [45]
It's B, 20in^2


So yeah!
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