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skelet666 [1.2K]
2 years ago
15

There are 20 students in my class. Each student needs: .2 cups, 2 plates, 1 party hat and a pair of earphones. My teacher has 2

boxes with plates and forks. Each box contains 8 plates and 10 forks. Would this be enough for the 20 students? Account for each plate and cup. Explain your answer.​
Mathematics
1 answer:
garik1379 [7]2 years ago
8 0

Answer:

Simple, no. This would not be enough.

Step-by-step explanation:

This is because you mentioned that there are 20 students and each of them needed 2 plates. So, there would need to be at least 40 plates. Forks seem irrelevant in this question but if the teacher has 8 plates in 1 box and another 8 in the second box, that would sum up to 16 plates that are available. And the fact that the box doesn't even include the other items, should hint the lack of items available for the students.

This question seemed worded differently. But tried my best. :)

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The solution is 4²=16, the are of circle A is sixteen times the area of circle B.
4 0
2 years ago
If you add six to twice a number, the result is sixteen. what is the number?
boyakko [2]

Answer: 5

Step-by-step explanation: The result, you are told is 16 equals 16 and that will then solve and find the value of X. So two X will be 10. Both sides take away six And then divide by two on both sides and thanks will be five.

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2 years ago
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Sav [38]

Answer:

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 7.5, \sigma = 1.1

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So

X = 8.6

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

Z = \frac{8.6 - 7.5}{1.1}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 6.4

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

Z = \frac{6.4 - 7.5}{1.1}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

6 0
3 years ago
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melamori03 [73]

Answer:

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4 0
2 years ago
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