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ANTONII [103]
2 years ago
5

Joseph claims that a scatterplot in which the y-values increase as the x-values increase must have a linear association. Amy cla

ims that the scatterplot could have a nonlinear association. Which statement about their claims is true?
Joseph is correct because only a line will increase along the whole data set. The scatterplot will have a positive, linear association.
Joseph is correct because only a line will decrease along the whole data set. The scatterplot will have a negative, linear association.
Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.
Amy is correct because only a nonlinear association could increase along the whole data set. A line has the same slope at any point, but a curve can get steeper at different points.
Mathematics
2 answers:
AleksandrR [38]2 years ago
7 0

Answer:

Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.

Step-by-step explanation:

Both linear and nonlinear associations could increase along the whole data set. That's why Joseph and the fourth option are incorrect.

Veronika [31]2 years ago
3 0

Answer:

The answer is C

Step-by-step explanation:

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P = 2(L + W)
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L = 4W - 7

136 = 2(4W - 7 + W)
136 = 2(5W - 7)
136 = 10W - 14
136 + 14 = 10W
150 = 10W
150/10 = W
15 = W <=== width is 15 m

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L = 4(15) - 7
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Answer:

<u>Question 1:</u>

<u>(a) </u>P(x<60) = 0.9236

<u>(b) </u>P(x>16) = 0.9564

<u>(c) </u>P(16<x<60) = 0.88

<u>(d) </u>P (x>60) = 0.0764

<u />

<u>Question 2:</u>

<u>(a) </u>P(x<3) = 0.0668

<u>(b) </u>P(x>7) = 0.0062

<u>(c) </u>P(3<x<7) = 0.927

<u />

Step-by-step explanation:

<u>Question 1:</u>

x = no. of mg of porphyrin per deciliter of blood.

μ = 40

σ = 14

(a) We need to compute P(x<60). We need to find the z-score using the normal distribution formula:

z = (x - μ)/σ

P(x<60) = P((x - μ)/σ < (60 - 40)/14)

             = P(z < 20/14)

             = P(z<1.43)

Using the normal distribution probability table we can find the value of p at z=1.43.

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                 = 1 - P(z<-1.71)

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(c) P(16<x<60) = P((16-40)/14) < x < (60-40)/14)

                        = P(-1.71 < z < 1.43)

This probability can be calculated as: P(z<1.43) - P(z<-1.71)

                                         P(16<x<60) = 0.9236 - 0.0436

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(d) P(x>60) = 1 - P(x<60)

     we have calculated P(x<60) in part (a) so,

    P(x>60) = 1 - 0.9236

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<u>Question 2:</u>                              

μ = 4.5 mm

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In this question, we will again compute the z-scores and then find the probability from the normal distribution table.

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               = 1 - P(z<2.5)

               = 1 - 0.9938

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(c) P(3<x<7) = P(x<7) - P(x<3)

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    P(3<x<7) = 0.9938 - 0.0668

    P(3<x<7) = 0.927

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