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Lostsunrise [7]
4 years ago
11

How many times dose 63 go into 272? THANK YOU SO MUCH FOR ANSWERING,IF YOU DID!!!!!!!

Mathematics
1 answer:
Anika [276]4 years ago
7 0
Let's write this problem as an expression: 272 ÷ 63
Now, it is just division, and we can divide the two numbers to get 4.317...
However, the question only asks for how many times 63 goes into 272, so the answer would be that 63 goes into 272 four times. Hope this helps and have an awesome day!

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Step-by-step explanation:

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A bag of chips weighs 228 grams. If each chip weighs 0.12 grams, how many chips are in the bag?
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1900

Step-by-step explanation:

so lets let x represent the total chips that in the bag.

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6 0
2 years ago
14 A cart loaded with 12 boxes weighs a total of 200
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8 0
3 years ago
The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a stan
IgorLugansk [536]

Answer:

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

Step-by-step explanation:

To solve this question, we need 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 3.02, \sigma = 0.29, n = 20, s = \frac{0.29}{\sqrt{20}} = 0.0648

Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

This is 1 subtracted by the palue of Z when X = 3.10. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 3.02}{0.0648}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907

1 - 0.8907 = 0.1093

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

5 0
4 years ago
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