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kati45 [8]
3 years ago
8

The perimeter of an equilateral triangle is 6 cm more than the perimeter of a square. The length of a side of the square is 3 cm

less than the length of a side of the equilateral triangle. Find the area of the square.
Mathematics
1 answer:
yan [13]3 years ago
5 0
3t=4s+5          (s=side of square, t= side of triangle)

s=t-3, t=s+3, using this value for t in the first equation gives us:

3(s+3)=4s+5

3s+9=4s+5

-s=-4

s=4

so the side length of the square is 4cm and the area of a square is just s^2

A=16cm^2
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Hello! Can someone help with this question?
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Hello! :)

\large\boxed{\text{70 seconds}}

Use a proportion to find how many seconds are needed to fold 5 shirts.

We are given that he takes 28 seconds to fold 2 shirts:

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Express in a fraction:

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3 years ago
Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an
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The probability that the mean life expectancy of the sample is less than X years is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is the mean life expectancy, \sigma is the standard deviation and n is the size of the sample.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

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In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

We have:

Mean \mu, standard deviation \sigma.

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

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True or False ?
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True the answer is true babes :)
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