The amount of space not occupied by the rubber balls is given by:
Volume=(volume of the container)-(volume of the rubber balls)
volume of the container is given by:
V=πr²h
V=π*(5/2)²(20)
V=392.70 cm³
Volume of each ball is:
V=4/3πr³
V=4/3π(2.5)³=65.45 cm³
volume of four balls
65.45×4=261.8 cm³
The volume of the container that is not occupied by the balls will be:
V=392.70-261.8
V=130.9 cm³
Answer:
72 cm³
Step-by-step explanation:
i just took the test :)
Answer:
A chip of 16.59 ounces represents the 67th percentile for the bag weight distribution
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 question, we have that:
![\mu = 16.5, \sigma = 0.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%2016.5%2C%20%5Csigma%20%3D%200.2)
What chip amount represents the 67th percentile for the bag weight distribution
This is X when Z has a pvalue of 0.67, so X when Z = 0.44.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.44 = \frac{X - 16.5}{0.2}](https://tex.z-dn.net/?f=0.44%20%3D%20%5Cfrac%7BX%20-%2016.5%7D%7B0.2%7D)
![X - 16.5 = 0.44*0.2](https://tex.z-dn.net/?f=X%20-%2016.5%20%3D%200.44%2A0.2)
![X = 16.59](https://tex.z-dn.net/?f=X%20%3D%2016.59)
A chip of 16.59 ounces represents the 67th percentile for the bag weight distribution