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lapo4ka [179]
3 years ago
8

Please help i will give brainiest!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
Lorico [155]3 years ago
4 0

Answer:

Part A: Two pieces of information that are provided by the graph is the median which is 55 and the range which is 20-95. Two piece of information that is not provided by the graph is the mean and mode.

Part B: The interquartile range of the data is 25. It represent the range between the amount of time in minutes that the students of a class surfed the internet.

I am not sure if this is all correct, but i tired. I hope this helps :)

FrozenT [24]3 years ago
3 0

Answer:

Part A: 2 pieces of information that are provided by the graph would be the median (median=55) and the range (which is 20-95) The 2 pieces of information that aren't provided by the graph are the mean and mode.

Part B: The IQR of the data is 25. It is the range between the amount of time in minutes that the students of a class surfed the internet.

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How do i write 2 perfect squares that each have a value greater then 100 and less that 200
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Construct the confidence interval for the population standard deviation for the given values. Round your answers to one decimal
Alex777 [14]

Answer:

The correct answer is "2.633< \sigma < 4.480".

Step-by-step explanation:

Given:

n = 21

s = 3.3

c = 0.9

now,

df = n-1

    =20

⇒ x^2_{\frac{\alpha}{2}, n-1 } = x^2_{\frac{0.9}{2}, 21-1 }

                  = 31.410

⇒ x^2_{1-\frac{\alpha}{2}, n-1 } = 10.851

hence,

The 90% Confidence interval will be:

= \sqrt{\frac{(n-1)s^2}{x^2_{\frac{\alpha}{2}, n-1 }} } < \sigma < \sqrt{\frac{(n-1)s^2}{x^2_{1-\frac{\alpha}{2}, n-1 }}

= \sqrt{\frac{(21-1)3.3^2}{31.410} } < \sigma < \sqrt{\frac{(21.1)3.3^2}{10.851} }

= \sqrt{\frac{20\times 3.3^2}{31.410} } < \sigma < \sqrt{\frac{20\times 3.3^2}{10.851} }

= 2.633< \sigma < 4.480

4 0
3 years ago
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