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juin [17]
1 year ago
8

What is the length of the apothem, rounded to the nearest inch? recall that in a regular hexagon, the length of the radius is eq

ual to the length of each side of the hexagon. 4 in. 5 in. 9 in. 11 in.
Mathematics
2 answers:
NNADVOKAT [17]1 year ago
8 0

Answer:

Step-by-step explanation:

What is the length of the apothem, rounded to the nearest inch? recall that in a regular hexagon, the length of the radius is equal to the length of each side of the hexagon. 4 in. 5 in. 9 in. 11 in.

Liula [17]1 year ago
8 0

Answer:

9

Step-by-step explanation:

apothem (AB)

1/2 of the length (10) is 5

AB^2 + 5^2 = 10^

AB^2 = 100-25

Square root of AB^2 = square root of 75

AB = 8.66

8.66 is about 9

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Answer:

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Step-by-step explanation:

8 0
3 years ago
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kap26 [50]
I don’t understand the question so I answer in two ways

4 0
2 years ago
A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

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2 years ago
Which space figure has six faces that are all regular quadrilaterals
nadya68 [22]
It would be a cube or a square prism. 
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3 years ago
Which number is a solution of the inequality x &lt;-4? Use the number line to help answer the question.
Fantom [35]

Answer:

0 is the answer

Step-by-step explanation:

4 0
3 years ago
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