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luda_lava [24]
2 years ago
7

Given: circle k(O), m LM = 164°, m WK = 68°, m∠MLK = 65° . Find: m∠LMW

Mathematics
1 answer:
uranmaximum [27]2 years ago
7 0

Answer:

The measure of angle LMW is

m\angle LMW=67\°

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of arc MW

we know that

The inscribed angle measures half that of the arc comprising

so

m\angle MLK=\frac{1}{2}[arc\ MW+arc\ WK]

substitute the given values

65\°=\frac{1}{2}[arc\ MW+68\°]\\ 130\°=[arc\ MW+68\°]\\ arc\ MW=130\°-68\°=62\°

step 2

Find the measure of arc LK

we know that

arc\ LM+arc\ MW+arc\ WK+arc\ LK=360\° -----> by complete circle

substitute the given values

164\°+62\°+68\°+arc\ LK=360\°\\ 294\°+arc\ LK=360\°\\ arc\ LK=360\°-294\°=66\°

step 3

Find the measure of angle LMW

we know that

The inscribed angle measures half that of the arc comprising

so

m\angle LMW=\frac{1}{2}[arc\ LK+arc\ WK]

substitute the given values

m\angle LMW=\frac{1}{2}[66\°+68\°]=67\°

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How may times will the number 3 be added?

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The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction
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Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.

Assuming that the random variable <em>X </em>follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

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  • For a sample size of <em>n</em> = 50

P(\mu-200

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  • For a sample size of <em>n</em> = 500

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<em>                                  n</em> :      20           50          100         500

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(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample (\bar x) approaches the whole population mean (\mu_{x}).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

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