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lions [1.4K]
3 years ago
12

A researcher is estimating the mean income of residents in a large city. The income variable is usually skewed to the right. She

collects a random sample of 25 people. The resulting 95% confidence interval is ($26700, $35400). Which one of the following conclusions is valid? We are 95% confident that the mean income for all residents of this city is between $26700 and $35400. 95% of the residents of this city have an income between $26700 and $35400. No conclusion can be drawn.
Mathematics
1 answer:
Gnom [1K]3 years ago
5 0

Answer: We are 95% confident that the mean income for all residents of this city is between $26700 and $35400.

Step-by-step explanation:

We know that a 95% confidence interval given an interval of values that we can be 95% sure , that it contains the true mean of the population, not 95% of data lies in it.

Given : A researcher is estimating the mean income of residents in a large city. The income variable is usually skewed to the right. She collects a random sample of 25 people.

The resulting 95% confidence interval is ($26700, $35400).

Then, valid conclusion will be :  We are 95% confident that the mean income for all residents of this city is between $26700 and $35400.

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Answer:

nonproportional

Step-by-step explanation:

x + 3

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3 years ago
Jonathan saved money to have a party if he spends to eight of his money on food 1/5 of his money on prices and two tenths of his
erik [133]

Answer:

7/20 is left

Step-by-step explanation:

2/8 (reduced is 1/4) for food

1/5 for prizes

2/10 (reduced is 1/5) for decorations

? for what's left

Add all the fractions up by finding a common denominator.

1/4 + 1/5 + 1/5

5/20 + 4/20 +4/20 = 13/20

He spent 13/20 of his money on this stuff. What part of that is left?

1 - 13/20 (AKA 20/20 - 13/20)

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Dominik [7]
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5 0
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Get this right and you get brainliest
blagie [28]

Answer:

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If my answer is incorrect, pls correct me!

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