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Komok [63]
3 years ago
9

The table shows data collected on the relationship between time spent playing video games and time spent with family. The line o

f best fit for the data is ý = -0.3694.5.
Video Games (Minutes) Time with Family (Minutes)
40 80
55 75
70 69
85 64
A) According to the line of best fit, the predicted number of minutes spent with family for someone who spent 36 minutes playing video games is 81.54.
B) Is it reasonable to use this line of best fit to make the above prediction?
1) The estimate, a predicted time of 81.54 minutes, is reliable but unreasonable.
2) The estimate, a predicted time of 81.54 minutes, is unreliable but reasonable.
3) The estimate, a predicted time of 81.54 minutes, is both unreliable and unreasonable.
4) The estimate, a predicted time of 81.54 minutes, is both reliable and reasonable.
Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0

Complete question :

The table shows data collected on the relationship between time spent playing video games and time spent with family. The line of best fit for the data is ý = -0.36x + 94.5.

Video Games (Minutes) Time with Family (Minutes)

40 80

55 75

70 69

85 64

A) According to the line of best fit, the predicted number of minutes spent with family for someone who spent 36 minutes playing video games is 81.54.

B) Is it reasonable to use this line of best fit to make the above prediction?

Answer:

81. 54

4) The estimate, a predicted time of 81.54 minutes, is both reliable and reasonable.

Step-by-step explanation:

Given that :

The equation of best fit is defined as :

ý = -0.36x + 94.5.

A.) Predicted number of minutes spent with family (y) for someone who spends 36 minutes playing games (x) will be :

x = 36

From the model :

ý = -0.36(36) + 94.5

y = - 12.96 + 94.5

y = 81.54

Hence.nunber of minutes spent with family = 81.54 minutes.

B.) The correlation Coefficient for the model is - 0.9993, which indicates a very strong negative relationship between both variables. Hence, using this fact, we can conclude that the prediction is reliable. From the data also, spending 36 minutes playing games means such individual will have more time than someone who plays games for 40 minutes, hence, the prediction is also reasonable.

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Step-by-step explanation:

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Hope this helped!

6 0
3 years ago
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The compressive strength of concrete is normally distributed with mu = 2500 psi and sigma = 50 psi. A random sample of n = 8 spe
Nata [24]

Answer:

X \sim N(2500,50)  

Where \mu=2500 and \sigma=50

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

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And the standard error would be:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Let X the random variable that represent the compressive strength of concrete of a population, and for this case we know the distribution for X is given by:

X \sim N(2500,50)  

Where \mu=2500 and \sigma=50

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

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Answer:

  x = -2.5 ± √14.25

Step-by-step explanation:

Add 5x to prepare for "completing the square."

  x² +5x = 8

Add the square of half the x-coefficient:

  x² +5x + 2.5² = 8 + 2.5²

  (x +2.5)² = 14.25

Take the square root.

  x +2.5 = ±√14.25

Subtract 2.5.

  x = -2.5 ± √14.25 ≈ {-6.27492, 1.27592}

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