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Lady bird [3.3K]
3 years ago
13

The ages of students in a school are normally distributed with a mean of 16 years and a standard deviation of 1 year. Using the

empirical rule, approximately what percent of the students are between 14 and 18 years old?
Mathematics
1 answer:
Dovator [93]3 years ago
7 0

Answer:

95% of students are between 14 and 18 years old

Step-by-step explanation:

First we calculate the Z-scores

We know the mean and the standard deviation.

The mean is:

\mu=16

The standard deviation is:

\sigma=1

The z-score formula is:

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

For x=14 the Z-score is:

Z_{14}=\frac{14-16}{1}=-2

For x=18 the Z-score is:

Z_{18}=\frac{18-16}{1}=2

Then we look for the percentage of the data that is between -2 deviations from the mean.

According to the empirical rule 95% of the data is less than 2 standard deviations of the mean.  This means that 95% of students are between 14 and 18 years old

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CaHeK987 [17]

Answer:

The coefficient of x is 8

Step-by-step explanation:

Given

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Required

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Collect like terms

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4 0
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The backyard is 20 inches long and 37 inches wide. What is the area of the backyard.
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            - - - - - - -- - - - - - - - - - - - - -- -

\diamond\large\textsf{\textbf{\underline{Given\:question:-}}}

          The backyard is 20 inches long and 37 inches wide. What is the area of the backyard?

\diamond\large\textsf{\textbf{\underline{\underline{Answer and how to solve:-}}}}

We are given the backyard's length (20 inches long) and the backyard's width (37 inches wide).

We are asked to find the area of the backyard, which can be found using the following formula:-

      \boxed{\bold{A_{(rectangle)}=LW}}

<u>Where:-</u>

<u></u>

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Substitute the values:-

\boxed{\bold{A=20*37}}

On simplification,

✳︎ \boxed{\bold{A=740\:in^2}}, which is our final answer.

<h3> Good luck with your studies.</h3>

        - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - -

5 0
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Step-by-step explanation:

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This is what it says
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