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Zielflug [23.3K]
3 years ago
9

There are 27 students in tanikas class. at lunch they sit at 3 tables if the same number of students sits at each table how many

students are at each table?
Mathematics
2 answers:
Troyanec [42]3 years ago
6 0
There are 9 at each table
Mashutka [201]3 years ago
4 0

Answer: 9

Step-by-step explanation:

Total number of Students =27

Total number of tables= 3

same number of students sit on a table

so,

27/3

9 students sit on a table

If you find this helpful mark me as brainlist.

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Line AB contains points A (4,5) and B(9,7). What is the slope of AB
ki77a [65]
This line crosses the point (4,5)so we can make an equation to find the slope:
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8 0
4 years ago
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The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated b
storchak [24]

Answer:

A task time of 177.125s qualify individuals for such training.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

A distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec, so \mu = 145, \sigma = 25.

The fastest 10% are to be given advanced training. What task times qualify individuals for such training?

This is the value of X when Z has a pvalue of 0.90.

Z has a pvalue of 0.90 when it is between 1.28 and 1.29. So we want to find X when Z = 1.285.

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

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X - 145 = 32.125

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A task time of 177.125s qualify individuals for such training.

7 0
4 years ago
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bearing in mind that an infinite geometric sequence, has a limit, namely converges at a value, only if "r" the common factor, is a proper fraction, namely | r | < 1, in this case it's so, thus


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