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4 years ago

Enrollment at ELAC decreased by 5%, or 600 people, the year. How many people were enrolled last year?

Mathematics

1 answer:

babymother [125]4 years ago

7 0

Answer:

12,000.

Step-by-step explanation:

Let x be the number of people that were enrolled last year.

We have been given that enrollment at ELAC decreased by 5%, or 600 people, the year. We are asked to find the number of people that were enrolled last year.

We can set as equation such that 5% of x equals 600.

$\frac{5}{100}\cdot x=600$

$0.05x=600$

$\frac{0.05x}{0.05}=\frac{600}{0.05}$

$x=12,000$

Therefore, 12,000 people were enrolled last year.

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Answer:

a) $1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75$

So we expected about 75% within two deviations from the mean

b) $1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556$

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

$Lower = 3.41 -1.5*0.09 = 3.275$

$Upper = 3.41 +1.5*0.09 = 3.545$

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

And using the lower limit we got:

$z = \frac{3.05-3.41}{0.09}=-4$

And with the upper limit we got:

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So then the value of k =4 and the percentage is given by:

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Step-by-step explanation:

Previous concepts and Data given

$\mu =3.41$ reprsent the population mean

$\sigma=0.09$ represent the population standard deviation

The Chebyshev's Theorem states that for any dataset

• We have at least 75% of all the data within two deviations from the mean.

• We have at least 88.9% of all the data within three deviations from the mean.

• We have at least 93.8% of all the data within four deviations from the mean.

Or in general words "For any set of data (either population or sample) and for any constant k greater than 1, the proportion of the data that must lie within k standard deviations on either side of the mean is at least: $1-\frac{1}{k^2}$

Part a

For this case we can find the percentage required replaincg k =2 and we got:

$1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75$

So we expected about 75% within two deviations from the mean

Part b

For this case we can find the percentage required replaincg k =2 and we got:

$1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556$

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

$Lower = 3.41 -1.5*0.09 = 3.275$

$Upper = 3.41 +1.5*0.09 = 3.545$

Part c

We can calculate how many deviations we are within the mean with the limits with this formula:

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

And using the lower limit we got:

$z = \frac{3.05-3.41}{0.09}=-4$

And with the upper limit we got:

$z = \frac{3.77-3.41}{0.09}=4$

So then the value of k =4 and the percentage is given by:

$1-\frac{1}{k^2} =1- \frac{1}{4^2}= 1-0.0625 = 0.9375$

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