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bija089 [108]
3 years ago
13

What's the area of a sector of a circle that has a diameter of 10 in. if the length of the arc is 10 in.?

Mathematics
2 answers:
Natali5045456 [20]3 years ago
8 0

Answer:

25 SQ IN. FOR PENN

Step-by-step explanation:


jasenka [17]3 years ago
5 0
"The diameter is 10, therefore radius = 5. If arc length = 10, then angle = 10/5 = 2 radians 2 radians = 2/2pi = 1/pi of a complete circle. 
Area of circle = pi * r^2 = 25pi. 
Area of sector = angle * area of circle = 1/pi * 25pi = 25"
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1Y, the number of accidents per year at a given intersection, is assumed to have a Poisson distribution. Over the past few years
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Answer:

The probability that the intersection will come under the emergency program is 0.1587.

Step-by-step explanation:

Lets divide the problem in months rather than in years, because it is more suitable to divide the period to make a better approximation. If there were 36 accidents in average per year, then there should be 3 accidents per month in average. We can give for the amount of accidents each month a Possion distribution with mean 3 and variance 3.

Since we want to observe what happen in a period of one year, we will use a sample of 12 months and we will take its mean. We need, in average, more than 45/12 = 3.75 accidents per month to confirm that the intersection will come under the emergency program.

For the central Limit theorem, the sample mean will have a distribution Normal with mean 3 and variance 3/12 = 0.25; thus its standard deviation is √0.25 = 1/2.

Lets call the sample mean distribution X. We can standarize X obtaining a standard Normal random variable W with distribution N(0,1).

W = \frac{X-\mu}{\sigma} = \frac{X-3}{1/2} = 2x-6

The values of \phi , the cummulative distribution function of W, can be found in the attached file. We are now ready to compute the probability of X being greater than 3.75, or equivalently, the probability than in a given year the amount of accidents is greater than 45, leading the intersection into an emergency program

P(X > 3.75) = P(2X-6 > 2*3.75-6) = P(W > 1) = 1-\phi(1) = 1-0.8413 \\= 0.1587

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