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2 years ago

Please help! thank you :)

Mathematics

1 answer:

omeli [17]2 years ago

4 0

Hi I’m not the best at math
But I think the answer is -4.2

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Of the 25 sheep in a flock 34% are white. how many sheep are white

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8.5 sheep are white. 34% of 25 is 8.5.

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What is the most preside classification of the quadrilateral formed by connecting the midpoints of the sides of a square?

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A professor grades students on four tests, a term paper, and a final examination. Each test counts as 15% of the course grade. T

Dmitry [639]

Answer:

Marilee wants to earn an "A" in a class and needs an overall average of at least Her test grades are 88 , 92,100 , and 80 . The average of her quizzes is 90 and counts as one test grade. The final exam counts as 2.5 test grades. What scores on the final exam would result in Marilee's overall average of 92 or greater? (See Example 9

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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:

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

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

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

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2 years ago

Nancy was trying to save up $457. At her job she made $8 an hour and she worked 40 hours a week. After paying for her food and o

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Thats 320 take home every week
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3 years ago