Weights measured in grams of randomly selected M&M plain candies:
1 answer:
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:
Sample standard deviation:
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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The equation of a parabola whose vertex is (0, 0) and focus is (1 / 8, 0) is equal to x = 2 · y².
<h3>How to derive the equation of the parabola from the locations of the vertex and focus</h3>
Herein we have the case of a parabola whose axis of symmetry is parallel to the x-axis. The <em>standard</em> form of the equation of this parabola is shown below:
(x - h) = [1 / (4 · p)] · (y - k)² (1)
Where:
- (h, k) - Coordinates of the vertex.
- p - Distance from the vertex to the focus.
The distance from the vertex to the focus is 1 / 8. If we know that the location of the vertex is (0, 0), then the <em>standard</em> form of the equation of the parabola is:
x = 2 · y² (1)
The equation of a parabola whose vertex is (0, 0) and focus is (1 / 8, 0) is equal to x = 2 · y².
To learn more on parabolae: brainly.com/question/4074088
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