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3 years ago

Find the area of the polygon.

Mathematics

2 answers:

professor190 [17]3 years ago

6 0

The formula for the area of the triangles is base x hight divided by 2 (bh/2)
The formula for the square is length times width (l x w)

therefore...

bh/2 = (12)(6) / 2 = 36 (this is the area of one triangle)

then you need to times 36 by 4 because you have 4 triangles.
36 x 4 = 144

l x w = 12 x 12 = 144 (this is the area of the square)

last you need to add the area of the triangles (144) and the area of the square (144) because that will give you the total surface area of the shape.

144+144= 288

so your answer is 288cm squared (288cm2)

JulsSmile [24]3 years ago

3 0

Answer:

its 216

Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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4 years ago

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4 years ago

Assume that the one-way commute time of an UoU student from his house to school is a normally distributed random variable which

Marizza181 [45]

Answer:

$n=(\frac{1.960(10)}{5})^2 =15.36 \approx 16$

So the answer for this case would be n=16 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

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.

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=10$ represent the sample standard deviation

n represent the sample size

ME=5 represent the margin of error

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (2)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (3)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.96$ , replacing into formula (3) we got: