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Nataliya [291]
3 years ago
12

What is the least possible sum of the digits displaying the time on a 12-hour digital clock?

Mathematics
1 answer:
Arlecino [84]3 years ago
4 0
<span>So if you think about it, this problem isn't that hard.So say it's 12:59, the total sum of those digits would be 71. if it was 1:11, then the sum would be 3. if it would be 3:47, the sum would be 50. So now, you have to find the time where it would be the least possible sum. That time would be 1:00 because 1+0+0=1.
</span><span> </span>
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(always remember to sort your negatives!)

Add like terms 

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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
UkoKoshka [18]

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:

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And we have;

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

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

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