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ehidna [41]
3 years ago
10

Is my work correct?

Mathematics
2 answers:
kow [346]3 years ago
7 0

Solution:

Given: A B CD is an inscribed polygon.

To Prove: ∠A and​ ∠C ​ are supplementary angles.

Proof: Join AC and B D.

Angle in the same segment of a circle are equal.

∠ACB=∠ADB→→AB is a segment.

Also, ∠A B D=∠A CD→→AD is a Segment.

In Δ ABD

∠A+∠ABD+∠ADB=180°→→Angle sum property of triangle.

∠A+∠A CD+ ∠ACB=180°

∠A+∠C=180°

Hence proved, that is, ∠A and​ ∠C ​ are supplementary angles.

The method Adopted by you

∠1=2 ∠A----(1)

and, ∠2=2 ∠C-------(2)

The theorem which has been used to prove 1 and 2, Angle subtended by an arc at the center is twice the angle subtended by it any point on the circle.→(Inscribed angle theorem)

Also, angle in a complete circle measures 360°.→→Chord arc theorem

∠1+∠2=360°→→Addition Property of Equality

2∠A+2∠C=360°→→[Using 1 and 2, Called Substitution Property]

Dividing both sides by 2→→Division Property of Equality

2∠A+2∠C=360°→→[Using 1 and 2]

∠A+∠C=180°

→→Correct work.

dsp733 years ago
4 0
Check the picture below.

so, notice, the intercepted arc by the inscribed angle A is the red arc there, namely arcBCD, and the one intercepted by angle C is the arcBAD in blue there.

now, both arcs take up the whole circle, namely the whole 360°, and we also know that arcBCD = 2A, whilst arcBAD = 2C.

\bf \stackrel{2(\measuredangle A)}{\widehat{BCD}}+\stackrel{2(\measuredangle C)}{\widehat{BAD}}=360\implies 2A+2C=360\implies 2(A+C)=360
\\\\\\
A+C=\cfrac{360}{2}\implies A+C=180

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a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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:

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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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 11, \sigma = 3

a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 25, s = \frac{3}{\sqrt{25}} = 0.6

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

Z = \frac{11.2 - 11}{0.6}

Z = 0.33

Z = 0.33 has a pvalue of 0.6293.

X = 10.8

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

Z = \frac{10.8 - 11}{0.6}

Z = -0.33

Z = -0.33 has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.5 and 11 ​minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

Z = \frac{11 - 11}{0.6}

Z = 0

Z = 0 has a pvalue of 0.5.

X = 10.5

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

Z = \frac{10.5 - 11}{0.6}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 100, s = \frac{3}{\sqrt{100}} = 0.3

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

Z = \frac{11.2 - 11}{0.3}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486.

X = 10.8

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

Z = \frac{10.8 - 11}{0.3}

Z = -0.67

Z = -0.67 has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

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