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astra-53 [7]
3 years ago
13

A certain element has a half-life of 69 years. An experiment starts with 1,500 grams of the element. The amount of the element r

emaining x years after the experiment began can be modeled with the function f(x) = 1,500(2)–x/69. The mathematical domain of the function is all real numbers.
Which statement describes how the reasonable domain compares to the mathematical domain?

1The reasonable domain is restricted to integers.
2The reasonable domain is restricted to positive real numbers.
3The reasonable domain has a minimum value of 1,500.
4The reasonable domain has a maximum value of 1,500.
Mathematics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

2. The reasonable domain is restricted to positive real numbers.

Step-by-step explanation:

The variable "x" is time in years from when the experiment began. It makes no sense to have negative values of x, as the experiment had not yet begun in negative time.

___

I would include x=0 in the domain, too, though you already know the amount of remaining element at x=0 and don't have to use the function to calculate it.

__

x might be restricted to integers if you're only measuring the remaining amount once a year.

The maximum value of 1500 applies to the *range* of the function, not its domain.

The minimum value of 1500 has nothing to do with anything.

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Step-by-step explanation:

62° + 90° = 152°

180° - 152° = 28°

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I need some help... HOW DO I DO THISSSS. NOT ASKING YOU TO DO IT JUST HELP ME
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6 0
3 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0
3 years ago
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