150 increased by 170% is what

1 answer:

7 0

To find any percentage increase, you multiply the decimal version of that perfect by the amount you want to find the percentage of, and the ADD THAT to your original price

formula: original price + original price (decimal version of the the percent)

so let's do that!

original price = 150

that's pretty easy to see, but what about the decimal version of the percent? all you gotta do is more the decimal to the left two times.

170% = 1.70

soooo decimal version of percent = 1.7

let's fill this into the formula!

150 + 1.7 (150) = 405 <<there's you're answer. hope that helps!

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Step-by-step explanation:

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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;

$\mu_{\bar X}= 14.70$

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

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