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3 years ago

In the diagram, m CD = 128° and m DA = 76°. What is m∠ABC?

2 answers:

7 0

Angle ABC is an inscribed angle.

Inscribed angle = 1/2 * Intercepted Arc

The intercepted arc is arc CD + arc DA, which would be 128 + 76 = 204

m<ABC = 1/2(204)

m<ABC = 102 degrees

You answer is D) 102

Vikki [24]3 years ago

6 0

Answer: The value of m∠ABC is 102° .

Step-by-step explanation:

Since we have given that

m CD = 128°

m DA = 76°

We need to find the m∠ABC, Angle subtended at the center is half of the sum of the measure of the arc intercepted by it.

$\frac{m\ CD+m\ DA}{2}=\angle ABC\\\\\frac{128^\circ+76^\circ}{2}=\angle ABC\\\\102^\circ=m\angle ABC$

The value of m∠ABC is 102° .

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In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

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Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

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