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ExtremeBDS [4]
2 years ago
6

. A restaurant charges a $100 setup fee, plus $15 per guest, to host a private party. The table

Mathematics
2 answers:
Aleks [24]2 years ago
5 0

Answer:

the CORRECT answer is $775

Step-by-step explanation:

multiply the guests by the amount for each then add the setup price.

45 x 15 = 675 + 100 = $775

guests=45

$ per guest=$15

setup price=$100

netineya [11]2 years ago
3 0

Answer:

42

Step-by-step explanation:

i said so

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The error.....Kelvin just figured the 60% markdown cost...not the final price

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or another way to do this is : if it is marked down 60%, then u r actually paying 40%......0.40(30.65) = 12.26
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Find the length of the longest line segment that can be drawn on a rectangular board 3.07m by 2.24m
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l=3.07m
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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil
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Answer:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

Step-by-step explanation:

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".  

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

X \sim N(12,0.3)  

Where \mu=12 and \sigma=0.3

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}

If we solve for the total T we got:

T= n \bar X

For this case then the expected value and variance are given by:

E(T) = n E(\bar X) =n \mu

Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2

And the deviation is just:

Sd(T) = \sqrt{n} \sigma

So then the distribution for the total would be also normal and given by:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

And we want this probability:

P(T\leq 72)

And we can use the z score formula given by:

z = \frac{x-\mu}{\sigma}

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

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