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alukav5142 [94]
3 years ago
15

For brainiest:):):):):):):):)

Mathematics
1 answer:
mars1129 [50]3 years ago
8 0

Answer:

A1) 125

A2) 47.5%

B) 30

Step-by-step explanation:

Remember 'of' in this sentence means times(*). % shows a certain number/100.

a1) 48% 'of' what number is 60?

(let unknown number be x)

48/100 *x=60

Solve the equation, and the answer is x=125

a2) what percentage 'of' 120 is 57?

(let the unknown percentage be x)

(x/100)*120=57

x=47.5%

b) 24 students= 80% of the total number of students

1%=24/80

100% of the total number of students=30

So, there are total of 30 students

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The graph of F(x) shown below resembles the graph of G(x) = x ^ 2 but it has been changed somewhat . Which of the following coul
sukhopar [10]

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f(x) = 3x^{2} + 2

Step-by-step explanation:

Given G(x) = x^{2}

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3 years ago
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:

\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

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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Since there is two negative signs in the original equation, you need two negative signs in the equivalent one as well.

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