To inscribe a hexagon inside a circle, to which length should you set your compass in order to draw the six vertices?

Mathematics

1 answer:

vesna_86 [32]4 years ago

6 0

Answer:

You need to set your compass to the radius of the circle.

Step-by-step explanation:

So therefore the answer is A)

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Can u please help me with this math assignment

Troyanec [42]

Hard to read sideways but what I got from that is that you have a right triangle and you are trying to explain the other angles?
Both of the other two angles would be 45 because an angle is 180 total. If it is a right angle, that means that there is a 90 degree angle. 180-90=90. In a right angle, the other two angles are the same in degree. Therefor 90/2 is 45. 45 applies to both of the acute angles. I hope this helps! Good luck :)

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3 years ago

Consider the probability that greater than 94 out of 153 people will not get the flu this winter. Assume the probability that a

Hatshy [7]

Answer:

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Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .