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nadezda [96]
2 years ago
5

Weights measured in grams of randomly selected M&M plain candies:

Mathematics
1 answer:
charle [14.2K]2 years ago
6 0

Answer:

The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Sample mean:

\overline{x} = \frac{0.957 + 0.912 + 0.925 + 0.886 + 0.920 + 0.958 + 0.915 + 0.914 + 0.947 + 0.939 + 0.842}{11} =  0.9195

Sample standard deviation:

s = \sqrt{\frac{(0.957-0.9195)^2 + (0.912-0.9195)^2 + (0.925-0.9195)^2 + (0.886-0.9195)^2 + ...}{10}} = 0.0336

What is the range of weights of the middle 99.7% of M&M’s?

By the Empirical Rule, within 3 standard deviations of the mean, so:

0.9195 - 3*0.0336 = 0.8187.

0.9195 + 3*0.0336 = 1.0203.

The range of weights of the middle 99.7% of M&M’s is between 0.8187 and 1.0203.

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Answer:

No, since 832 ≠ 900, using Pythagorean theorem, the width should be 18 in

Step-by-step explanation:

It is better to type the question in the 'Question' than the fact that you need help so that we can start seeing the problem immediately.

A rectangle has 4 right angles so if l = 24 and w = 16 then you can use the pythagorean theorem to see if this is possible since

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The solutions of the equations are the point of intersection of the equations

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<h3>How to determine the solutions?</h3>

The equations are given as:

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