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ozzi
3 years ago
9

Is it a square, rhombus, or a rectangle? Why?

Mathematics
2 answers:
katrin [286]3 years ago
8 0

Answer:

Technically, this figure is all of them, but more specifically, a square

Step-by-step explanation:

Definitions:

Rectangle: Quadrilateral that is equiangular

Rhombus: Quadrilateral that is equilateral

Square: An equiangular, equilateral quadrilateral

Counting the sides of all sides show us that <em>this quadrilateral is equilateral</em>, and knowing that all x = ₙ and y = ₙ are perpendicular shows us that<em> this quadrilateral is equiangular</em>.

So, this figure is all of those shapes, most specifically a square.

Alex73 [517]3 years ago
7 0

Answer:

It is a square.

Step-by-step explanation:

If you look at the space of the lines from point to point, it's a even amount of little squares, so therefore, it is a square.

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In a given year, the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500. If a samp
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15.15% probability that the sample mean will be $192,000 or more.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 189000, \sigma = 20500, n = 50, s = \frac{20500}{\sqrt{50}} = 2899.14

The probability that the sample mean will be $192,000 or more is

This is 1 subtracted by the pvalue of z when X = 192000. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{192000 - 189000}{2899.14}

Z = 1.03

Z = 1.03 has a pvalue of 0.8485.

1-0.8485 = 0.1515

15.15% probability that the sample mean will be $192,000 or more.

7 0
2 years ago
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