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NeX [460]
3 years ago
6

The length of a rectangle is 2 units greater than the width. The area of the rectangle is 24 square units. What is it's width?

Mathematics
2 answers:
eimsori [14]3 years ago
8 0

Answer:

Width = 4 units

Step-by-step explanation:

Let us pose the width as w, and the length as l. If the length is 2 units greater than the width, consider the following;

l = 2 + w,\\\\w = width,\\l = length

The area of this rectangle can be determined through length * width / l * w, and is given to be 24 square units. We can say l = 2 + w instead, solving for the width ( w );

( 2 + w ) * w = 24,\\2w+w^2=24,\\\left(w-4\right)\left(w+6\right)=0,\\w = 4, w = - 6\\\\Solution - width = 4 units

As the width couldn't be a negative value, we had to take the positive of 4 and - 6, which was 4 units.

gogolik [260]3 years ago
7 0

Answer:

4 units

Step-by-step explanation:

Length would be 6 units and width would be 4 units

6*4 = 24 proving my claim to be true

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Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

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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 = 9.6, \sigma = 0.5.

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when X = 10.

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

Z = \frac{10 - 9.6}{0.5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

Z = \frac{8 - 9.6}{0.5}

Z = -3.2

Z = -3.2 has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

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Now we have n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1.

This probability is 1 subtracted by the pvalue of Z when X = 9.8. So:

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

Z = \frac{9.8 - 9.6}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

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