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1 year ago

8

Which statements describe a residual plot for a line of best fit that is a good model for a scatterplot? Check all that apply. T

here are about the same number of points above the x-axis as below it. The points are randomly scattered with no clear pattern. The points lie on a line. The points lie on a curve. The number of points is equal to those in the scatterplot. The y-coordinates of the points are the same as the points in the scatterplot.

2 answers:

lidiya [134]1 year ago

8 0

Answer:

The points are randomly scattered with no clear pattern

The number of points is equal to those in the scatterplot.

Step-by-step explanation:

The points in the residual plot of the line of best fit that is a good model for a scatterplot are randomly scattered with no clear pattern (like a line or a curve).

The number of points in the residual plot is always equal to those in the scatterplot.

It doesn't matter if there are about the same number of points above the x-axis as below it, in the residual plot.

The y-coordinates of the points are not the same as the points in the scatterplot.

babymother [125]1 year ago

3 0

Answer:

its a,b,e on e2020

Step-by-step explanation:

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Adam buys a bigger turkey than he needs.

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He should start cooking at 15:00

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2 years ago

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During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

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1 year ago

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Answer:

Hope it is correct :) :)

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2 years ago

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I think it might be 27

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The yearly cost in dollars, y, for a member at a video game arcade based on total game tokens purchased, x, is y =y equals Start

const2013 [10]

Answer:

$y = \frac{1}{5}x$

Step-by-step explanation:

Given

Yearly Cost for members; $y = \frac{1}{10}x + 6$

Required

Determine the yearly cost for non members

From the question, we understand that:

<em>A non member pays $0.20 for each game;</em>

Yearly cost is then calculated as thus;

y = Amount * Number of Games

Where:

Amount = $0.2 and Number of Games = x

$y = 0.20 * x$

Convert 0.2 to fraction

$y = \frac{2}{10} * x$

Divide the numerator and the denominator by 2

$y = \frac{1}{5} * x$

$y = \frac{1}{5}x$

<em>Hence, the yearly cost for non members is </em>

$y = \frac{1}{5}x$

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