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ANTONII [103]
3 years ago
5

Need Help ASAP only correct answers. Working on it now. Took a picture. question: Figure ABCDE is the result of a 180° rotation

of figure LMNOP about the point. Which angle in the pre- image corresponds to <D in the image? Enter your answer in the box.​

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
7 0

Answer:∠M in the pre-image corresponds to ∠B in the image.

Step-by-step explanation:

When  we look at the figures , we can see that the three vertices B , C , D are one side as vertices to M , N  ,O .

Clearly , vertex C corresponds to vertex N ( presents in the middle )

Also, point B is at the right from point C .In  Figure L M N O P , vertex M is at the right from vertex N .Thus , vertex M corresponds the vertex B.

It means ∠M in the pre-image corresponds to ∠B in the image.

brain list this answer thankyou:)

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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
Len [333]

Answer:

(a) <em>                             </em><em>n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

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:

  • For a sample size of <em>n</em> = 20

P(\mu-200

                                           =P(-0.37

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

P(\mu-200

                                           =P(-0.59

  • For a sample size of <em>n</em> = 100

P(\mu-200

                                           =P(-0.83

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

P(\mu-200

                                           =P(-1.86

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

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(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.

Thus, the correct option is (b).

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