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oee [108]
2 years ago
11

What does Arc C D E describe? a minor arc a major arc a semicircle a chord

Mathematics
2 answers:
Vikentia [17]2 years ago
8 0

Answer:B major arc

Step-by-step explanation:

edg

lana [24]2 years ago
7 0

Answer:

Major Arc should be your answer

Trust

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A, c, and e are the correct answers
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Natalie almost always sleep in on Saturday mornings when she does not have to work. If it is Saturday morning and Natalie does n
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yes i would except her to sleep in. Because every saturday if she almost always. And to day is a saturday she would sleep in

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2 years ago
Justin ate 1/4 of a pie. If there were 8 slices of pie, how many sclices did justin eat
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Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to
attashe74 [19]

Answer:

0.9726 = 97.26% approximate probability that X is at most 30

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 distributions 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)}.

11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped).

This means that p = 0.11

Random sample of 200 shafts

This means that n = 200

Mean and Standard deviation:

\mu = E(x) = np = 200*0.11 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.11*0.89} = 4.42

(a) What is the (approximate) probability that X is at most 30

Using continuity correction, this is P(X \leq 30 + 0.5) = P(X \leq 30.5), which is the pvalue of Z when X = 30.5. So

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

Z = \frac{30.5 - 22}{4.42}

Z = 1.92

Z = 1.92 has a pvalue of 0.9726.

0.9726 = 97.26% approximate probability that X is at most 30

8 0
3 years ago
PLZ HELP IM SO STUCK
maw [93]
I'm not sure but, maybe if you divide 300000 by 5.5% and see what you get. Then times it by 6. Not sure if it's correct, maybe see someone else's answer?
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