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german
3 years ago
8

Riley buys office equipment with a total price of $6,280 for her new company. If the office equipment depreciates at a rate of 1

3% each
year, which equation represents the value of the equipment after x years?
Mathematics
1 answer:
alexdok [17]3 years ago
4 0

Answer: y = 6280(0.87)^x

Step-by-step explanation:

If the office equipment depreciates at a rate of 13% each year, then the rate of depreciation is exponential.

We would apply the formula for exponential decay which is expressed as

y = b(1 - r)^x

Where

y represents the value of the office equipment after t years.

x represents the number of years.

b represents the initial value of the office equipment.

r represents rate of decay.

From the information given,

y = $628

r = 13% = 13/100 = 0.13

Therefore, the equation that represents the value of the equipment after x years is

y = 6280(1 - 0.13)^x

y = 6280(0.87)^x

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Step-by-step explanation:

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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:

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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;

\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

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