Refer to the diagram shown below.
Assume that approximately all the data is contained under the probability distribution curve within 3 standard deviations from the mean. It is actually 99.7%.
Therefore,
bin #1 corresponds to μ-3σ,
bin #11 corresponds to μ,
bin #21 corresponds to μ + 3σ.
Let x be the random variable for bin 17.
Then by interpolation,
![\frac{x-\mu}{(\mu + 3\sigma )-\mu} = \frac{17-11}{21-11} \\\\ x-\mu = 1.8\sigma](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx-%5Cmu%7D%7B%28%5Cmu%20%2B%203%5Csigma%20%29-%5Cmu%7D%20%3D%20%5Cfrac%7B17-11%7D%7B21-11%7D%20%5C%5C%5C%5C%20x-%5Cmu%20%3D%201.8%5Csigma)
The z-score for x is
![z= \frac{x-\mu}{\sigma} = \frac{1.8\sigma}{\sigma} =1.8](https://tex.z-dn.net/?f=z%3D%20%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%20%3D%20%5Cfrac%7B1.8%5Csigma%7D%7B%5Csigma%7D%20%3D1.8)
The probability corresponding to bin #17 is (from standard tables)
P(z≤1.8) = 0.964
Therefore if there are 1000 balls, the sum of the balls in bins 1 to 17 is
0.964 * 1000 = 96.4 ≠ 96 balls.
Answer: 96 balls
That’s the simplest form of 6/1 but some equivalent fractions would be, 12/2 = 18/3 = 24/4 and more
300,000+40,000+5,000
Three hundred forty-five thousand
First, we know that the cubes are identical, therefore, they would have half the combined volume.
Each cubes's individual volume is 250/2 = 125 cubic cm
Since we know that this shape is a cube, we also know that all the dimensions of a cube are equal.
v=s³ where v is the volume and s is the side length
125 cubic cm=s³
∛125=s
5 cm=s
One of the cube's edge dimensions would be 5 cm.
Hope I helped :)
Answer:
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Step-by-step explanation:
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