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3 years ago

What is the area of the figure below?

Mathematics

1 answer:

-BARSIC- [3]3 years ago

6 0

Answer:

54.8 = total area+

Step-by-step explanation:

Start with the widest part near the middle. Three inches is taken away from 7 inches to give 4 inches.

4 inches must be divided by 2 since the triangles so formed are congruent and there are 4 of them.

Use the Pythagorean theorem to get the height of the triangles. Solve the height for just one of them.

height = ??

Base = 2

hypotenuse = 5

height^2 + 2^2 = 5^2

height^2 + 4 = 25

height^2 + 4 - 4 = 25 - 4

height^2 = 21

height = sqrt(21)

height = 4.58

Area of 4 Trangles

Area of the triangles = 4 * (1/2 * 2 * 4.58)

Area of the triangles = 18.32

Area of the Middle Rectangle

Area = l * w

length = 4.58 * 2 where did that 2 come from?

width = 3

Area = 3 * 2 * 4.58

Area = 27.48

Area of the two End Squares

The end squares measure 3 by 3

Area square = s*s

The area = 3*3 = 9

Total Area

Area = 9 + 27.48 + 18.32

Area = 54.8 units.

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3 years ago

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3 years ago

The heights of baby giraffe are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 b

ANEK [815]

Answer:

$P(\bar X$

And we can solve this using the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we use this formula we got:

$z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4$

So we can find this probability equivalently like this:

$P( Z$

Step-by-step explanation:

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".

Solution to the problem

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

$X \sim N(63.6,2.5)$

Where $\mu=63.6$ and $\sigma=2.5$

We select n =100. Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We want this probability:

$P(\bar X$

And we can solve this using the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we use this formula we got:

$z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4$

So we can find this probability equivalently like this:

$P( Z$

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4 years ago

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Natasha2012 [34]

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sertanlavr [38]

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