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3 years ago

Each side of the regular hexagon below measures 8 cm. What is the area of the hexagon?

Mathematics

2 answers:

Rasek [7]3 years ago

8 0

Answer

Find out the area of the regular hexagon.

To prove

As given in the question

Hexagon is regular thus it is made up of the six equilateral triangle.

Thus area of the regular hexagon is 6 times the area of a equilateral triangle.

Each side of the regular hexagon measures 8 cm.

Formula

$Area\ of\ hexagon = 6\times \frac{\sqrt{3}a^{2}}{4}$

where a represented the side.

$Area\ of\ hexagon = 6\times \frac{\sqrt{3}\times 8\times 8}{4}$

$Area\ of\ hexagon = 6\times \sqrt{3}\times 8\times 2$

$Area\ of\ hexagon = 96 \sqrt{3}\ unit^{2}$

Therefore the option (a) is correct .

sineoko [7]3 years ago

6 0

A regular hexagon can be split into 6 congruent equilateral triangles.
the area of an equilateral triangle is √3*s²/4, where s is the side length
Area of the hexagon
= 6*√3*8²/4
= 96√3 square centimeters

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<h3>Explanation:</h3>

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Solve the equation for b-term.

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3 years ago

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2 years ago

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USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

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When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Gas mileage A: Mean 36, standard deviation 6, sample of 50:

$\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485$

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

$\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314$

Distribution of the difference:

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$\mu = \mu_A - \mu_B = 36 - 42 = -6$

Standard error:

$s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142$

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This is the difference of means, that is, -6 mpg.

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We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

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C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

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3 years ago