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Taya2010 [7]
3 years ago
13

The standard height from the floor to the bull’s-eye at which a standard dartboard is hung at 5 feet 8 inches. A standard dartbo

ard is 18 inches in diameter. Suppose a standard dartboard is hung at standard height so that the bull’s-eye is 10 feet from the wall to its left. Sasha throws a dart at the dartboard that land at point 10.25 Feet from the left wall and 5 feet above the floor. Does Sasha’s dart land on the dartboard? Drag the choices into the boxes to correctly complete the statements.

Mathematics
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

Hello! I'm sorry I couldn't get to your question sooner. I just completed this quiz!

The equation of the circle that represents the dartboard is (x-10)^2 + (y-17/3)^2 = 9/16, where the origin is the lower-left corner of the room and the unit of the radius is feet.  

The position of Sasha's dart is represented by the coordinates (10.25,5). Sash's dart does land on the dartboard.

This quiz was completed on k12, lesson 3.03.

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Answer:

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Step-by-step explanation:

3 0
2 years ago
For a given IQ test, an individual is considered a genius if their score falls more than three standard deviations from the mean
ki77a [65]

Answer:

P(Z>3) = 1-P(Z

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And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a.2

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu=?,\sigma=?)  

We are interested on this probability

P(X>X+3\mu)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

And we can find the following probablity:

P(Z>3) = 1-P(Z

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a/2.0

3 0
2 years ago
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PS

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2 years ago
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