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3 years ago

Write a rational function with no vertical asymptotes and no holes. Please explain.

Mathematics

1 answer:

Ksju [112]3 years ago

4 0

ANSWER

$f(x) = \frac{ {x}^{2} }{ {x}^{2} + 1 }$

EXPLANATION

We write the function such that both the numerator and the denominator are prime.

An example of a rational function with no vertical asymptotes and no holes is

$f(x) = \frac{ {x}^{2} }{ {x}^{2} + 1 }$

For the above rational function, the denominator is never zero, so there are no vertical asymptotes.

Also the highest common factor for the numerator and the denominator is 1 so there are no holes.

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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:

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.

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Answer:

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