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Sphinxa [80]
2 years ago
7

One-quarter of the counters in a tub are red. The rest of the counters are green. If there are 48 green counters, how many red c

ounters are there?
Mathematics
1 answer:
scZoUnD [109]2 years ago
6 0

There are 16 red counters there.

<h3>Fractions and proportions</h3>

If one-quarter of the counters in a tub are red and the rest are green, then there are three-quarter of the counter in the tub.

Let the total counter be x such that

3/4x = 48

3x = 4 * 48

x = 4 * 16

x = 64

Determine the amount of red counter

Red counter = 64 - 48

Red counter = 16

Hence there are 16 red counters there.

Learn more on proportion here; brainly.com/question/1496357

#SPJ1

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The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m
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Answer:

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we have;

\mu_{\bar X}= 14.70

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Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of \bar X Round your answers to two decimal places.

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".  

Solution to the problem

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

4 0
3 years ago
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