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tatiyna
3 years ago
9

A catering service offers 6 appetizers 7 main courses and 3 desserts A customer is to select 3 appetizers 3 main courses and 2 d

eserts for a banquet in how many ways can this be done

Mathematics
1 answer:
maria [59]3 years ago
3 0


3 appetizer then you mark down 3 main courses for each one the on each main dish add 2 desserts

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$12.50 divided by 3<br> i got 5 minutes to answer
Llana [10]
About $4.17 or $4.20

Hope this helps!

Comment if you have questions!
7 0
3 years ago
Eight hotdogs at the basketball park cost $28 how much does one hotdog cost
Igoryamba
28 divided by 8 is equal to 3.5, so one hotdog will cost $3.50
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3 years ago
7000000000000000000000 people ate 800 pizzas how many slices would each person get?
aleksklad [387]
Hello! I may be wrong, I’m just going off the calculator LOL. I got:

5.6e+24

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7 0
3 years ago
A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20 randomly selected adults are given an IQ t
Fiesta28 [93]

Answer:

100% probability that the sample mean scores will be between 87 and 124 points

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 105, \sigma = 20, n = 20, s = \frac{20}{\sqrt{20}} = 4.47

What is the probability that the sample mean scores will be between 87 and 124 points

This is the pvalue of Z when X = 124 subtracted by the pvalue of Z when X = 87. So

X = 124

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{124 - 105}{4.47}

Z = 4.25

Z = 4.25 has a pvalue of 1

X = 87

Z = \frac{X - \mu}{s}

Z = \frac{87 - 105}{4.47}

Z = -4.25

Z = -4.25 has a pvalue of 0

1 - 0 = 1

100% probability that the sample mean scores will be between 87 and 124 points

3 0
4 years ago
Whats the answer of<br> 2 1/10+7/10
Vsevolod [243]
2 1/10+7/10
=2 8/10
=2 4/5

3 0
3 years ago
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