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Bezzdna [24]
2 years ago
15

Please help I will give brainly to the one with the best answer

Mathematics
1 answer:
disa [49]2 years ago
3 0
I think you could try adding up all the x and then divided
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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
hoa [83]

Answer:

Obese people

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

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

Lean People

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

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

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".  

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

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.

5 0
3 years ago
A church group ordered a total of 22 drinks and burgers.each drink cost $2.50 and each burger cost $6.25 if the group spent a to
stepan [7]
14.8 or round up to 15 burgers
7 0
4 years ago
Kylie buys a ceramic case priced at $75. Shipping and handling are an additional 9 2/5% of the price. How much shopping and hand
salantis [7]
\bf \begin{array}{ccllll}
amount&\%\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
75&100\\
x&9\frac{2}{5}
\end{array}\implies \cfrac{75}{x}=\cfrac{100}{9\frac{2}{5}}\implies \cfrac{75}{x}=\cfrac{100}{\frac{9\cdot 5+2}{5}}

\bf \cfrac{75}{x}=\cfrac{\frac{100}{1}}{\frac{47}{5}}\implies \cfrac{75}{x}=\cfrac{100}{1}\cdot \cfrac{5}{47}\implies \cfrac{75}{x}=\cfrac{500}{47}\implies 75\cdot 47=500\cdot x
\\\\\\
\cfrac{3525}{500}=x\implies \cfrac{141}{20}=x\implies 7.05=x
7 0
3 years ago
I really need help ASAP!
Semmy [17]

Answer:

Because they are parallel:

Angles 1,3,5 and 7=125

180-125=55

Angles 2,4, 6 and 8=55

Hope this helps!

3 0
3 years ago
Does anyone know this
Tamiku [17]

Answer:

I belive the answer is A

Step-by-step explanation:

So any answer with 22t would make sense, so you have A and C. In C though, it is subtracting 22, but since 6195 is the total it would have to include the 22 so it is A.

8 0
3 years ago
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