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vova2212 [387]
3 years ago
11

Find the measure of ∠ECB. Circle A is intersected by line CD at points D and E and line CB at point B, forming angle ECB outside

of the circle, the measure of angle ECB is x plus 10 degrees, arc EB is 6x plus 6 degrees, and arc DB 146 degrees.
Mathematics
2 answers:
goldfiish [28.3K]3 years ago
6 0

Answer:

its 45

Step-by-step explanation:

lubasha [3.4K]3 years ago
3 0

Answer:

m\angle ECB=50^o

Step-by-step explanation:

we know that

The measure of the external angle is the semi-difference of the arches it covers

so

In this problem

m\angle ECB=\frac{1}{2}(arc\ EB-arc\ DB)

substitute the given values

(x+10)^o=\frac{1}{2}((6x+6)^o-146^o)

solve for x

2x+20=6x-140\\6x-2x=20+140\\4x=160\\x=40

Find the measure of ∠ECB

substitute the value of x

m\angle ECB=(40+10)=50^o

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A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

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.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

15 is the bounds with want, 25 is the standard error. So

Z = 15/25 = 0.6 has a pvalue of 0.7257

Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

45.14% probability that the point estimate was within ±15 of the population mean

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Answer:

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