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Novosadov [1.4K]
3 years ago
12

Enter an inequality that represents the description, and then solve. Toni can carry up to 24 lb in her backpack. Her lunch weigh

s 1 lb, her gym clothes weigh 3 lb, and her books (b) weigh 4 lb each. How many books can she carry in her backpack?
Mathematics
2 answers:
stellarik [79]3 years ago
3 0
4b+4=24 the answer is 5

deff fn [24]3 years ago
3 0
24 - ( 1+3 ) = 20. 
20 + b(book weight; unknown) = answer

Since 4 x 5 is 20, the answer is that Toni can carry 5 books.
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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
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Answer:

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Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

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Lower = \mu - 2\sigma = 526- 2(107) = 312

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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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

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Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

X \sim N(373,67)  

Where \mu=373 and \sigma=67

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

X \sim N(526,107)  

Where \mu=526 and \sigma=107

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

The interval for the lean people is significantly higher than the interval for the obese people.

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Step-by-step explanation:

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