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exis [7]
4 years ago
13

A rectangular exercise mat has a perimeter of 36 feet. The length of the mat is twice the width. Write and solve an equation to

determine the length, in feet, of the mat. Then find the area, in square feet, of the mat.
Mathematics
1 answer:
Karo-lina-s [1.5K]4 years ago
7 0
<h2>Hello!</h2>

The answers are:

- Length

length=2width=12feet

- Width

width=6feet

- Area

Area=72feet^{2}

<h2>Why?</h2>

We are given the perimeter of a rectangular mat, and we are asked to find the area, and write an equation to determine the length, if feet, of the mat. To solve this problem we need to remember the equation of the perimeter of a rectangle, which is:

Perimeter=2length+2width

Using the given information we have,

36feet=2length+2width

Also, we know that the length is twice the width, so:

length=2width

Then, substituting the second equation into the first equation, we can find the width of the mat.

Substituting, we have:

36feet=2(2width)+2width

36feet=4width+2width

36feet=6width

[tex]\frac{36feet}{6} =width\\width=\frac{36feet}{6} =6feet[/tex]

Then, substituting the width into the second equation, we can find the length of the mat,

Substituting, we have:

length=2width

length=2(6feet)=12feet

Then, finding the area, we have:

A=length*width=12feet*6feet=72feet^{2}

Have a nice day!

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IQ scores are measured with a test designed so that the mean is 116 and the standard deviation is 16. Consider the group of IQ s
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Answer:

So the z-scores that separate the unusual IQ scores from those that are​ usual are Z = -2 and Z = 2.

The IQ scores that separate the unusual IQ scores from those that are​ usual are 84 and 148.

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 question, we have that:

\mu = 116, \sigma = 16

What are the z scores that separate the unusual IQ scores from those that are​ usual?

If Z<-2 or Z > 2, the IQ score is unusual.

So the z-scores that separate the unusual IQ scores from those that are​ usual are Z = -2 and Z = 2.

What are the IQ scores that separate the unusual IQ scores from those that are​ usual?

Those IQ scores are X when Z = -2 and X when Z = 2. So

Z = -2

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

-2 = \frac{X - 116}{16}

X - 116 = -2*16

X = 84

Z = 2

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

2 = \frac{X - 116}{16}

X - 116 = 2*16

X = 148

The IQ scores that separate the unusual IQ scores from those that are​ usual are 84 and 148.

8 0
3 years ago
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