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3 years ago

Wha is the area of this

Mathematics

1 answer:

Korvikt [17]3 years ago

6 0

Octagon, stop sign.

Eight isoscles triangles. It looks like we're told the side is 9.9 and the height to the side (also called the apothem) is 12.

So each isosceles triangle has area (1/2)(9.9)(12) and we have eight of them,

area = 8(1/2)(9.9)(12) = 475.2

Answer: 475.2

Usually we wouldn't be told 9.9 -- this is the baby version. We know each of those isoscles triangles has unique angle 360/8=45 degrees, so the apothem and half the side of the octagon are a right triangle with acute angle 22.5 degrees.

The area of the right triangle with long leg 12, short leg x,

tan 22.5 = x/12 or

x = 12 tan 22.5

Twice that is what we're told is 9.9; let's check:

2x = 24 tan 22.5 = 9.941125496954282

The area of the little right triangle is

(1/2) 12 × 12 tan 22.5

and there are 16 of these

16 (1/2) 12 × 12 tan 22.5 ≈ 477.174

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Help needed ASAP will give BRAINLIEST not a real test

sergey [27]

C
You would do 12x30 for the 12 months in 30 years = 360 then you would do 643.33x360 and that’s your answer

4 0

3 years ago

MNM Corporation gives each of its 500 employees an aptitude test. The scores on the test are normally distributed with a mean of

djyliett [7]

Answer:

Q1: $P(\bar X$

Q2: $0.6 - 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.504$

$0.6 + 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.696$

And the 95% confidence interval would be given (0.504;0.696).

We are confident 95% that about 50.4% to 69.6% of people would like the cereal

Q3: $n=\frac{0.5(1-0.5)}{(\frac{0.09}{1.96})^2}=118.57$

And rounded up we have that n=119

Step-by-step explanation:

Q1 :MNM Corporation gives each of its 500 employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 36 is taken from the population of 500 employees. What is the probability that the average aptitude test score in the sample will be less than 78.69?

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".

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

$X \sim N(75,15)$

Where $\mu=75$ and $\sigma=15$

The distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=75, \frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{36}}=2.5)$

For this case we want this probability:

$P(\bar X$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar X}}$

$P(\bar X$

Q2: A new brand of breakfast cereal is being market tested. One hundred boxes of the cereal were given to consumers to try. The consumers were asked whether they liked or disliked the cereal. Their responses are provided below.

Liked =60, Disliked =40 , Total =100

Construct a 95% confidence interval for the true proportion of people who will like the cereal. Interpret the practical meaning of this interval estimate, in plain English.

n=100 represent the random sample taken

X=60 represent the people liked with the cereal

$\hat p=\frac{60}{100}=0.6$ estimated proportion of people who liked the cereal

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Confidence =95% or 0.95

p= population proportion of people drivers claimed they always buckle up

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.6 - 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.504$

$0.6 + 1.96 \sqrt{\frac{0.6(1-0.6)}{100}}=0.696$

And the 95% confidence interval would be given (0.504;0.696).

We are confident 95% that about 50.4% to 69.6% of people would like the cereal

Q3: Assume that nothing is known about the population proportion. With a .95 probability, how large of a sample needs to be taken to provide a margin of error of 9% or less?

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.09$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.5 since we don't have other info provided to assume a different value. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.09}{1.96})^2}=118.57$

And rounded up we have that n=119

5 0

3 years ago

Miss Martin went to the grocery store and spent 28.08 dollars on 12 bags of salt and vinegar chips if they each cost the same am

Ne4ueva [31]

Answer:

2.34

Step-by-step explanation:

Divide the total by the number of bags!

28.08/12=2.34

CHECK: 2.34*12=28.08

brainliest, please? ty! <3

7 0

3 years ago

Does anyone know what the answer please?

Tasya [4]

The answer is d........

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2 years ago

What is the value of e^Ln7x ? A. 1 B. 7e C. 7x D. 7

AURORKA [14]

E and ln are each other's opposite. So the answer is simply 7x

6 0

3 years ago

Read 2 more answers