Answer:
The 90th percentile of the distribution is 6.512 ml.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 6 milliliters (ml) and a standard deviation of 0.4 ml.
This means that ![\mu = 6, \sigma = 0.4](https://tex.z-dn.net/?f=%5Cmu%20%3D%206%2C%20%5Csigma%20%3D%200.4)
Find the dye amount that represents the 90th percentile (i.e. 90%) of the distribution.
This is X when Z has a p-value of 0.9, so X when Z = 1.28. Then
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.28 = \frac{X - 6}{0.4}](https://tex.z-dn.net/?f=1.28%20%3D%20%5Cfrac%7BX%20-%206%7D%7B0.4%7D)
![X - 6 = 1.28*0.4](https://tex.z-dn.net/?f=X%20-%206%20%3D%201.28%2A0.4)
![X = 6.512](https://tex.z-dn.net/?f=X%20%3D%206.512)
The 90th percentile of the distribution is 6.512 ml.
B
the volume (V) of a sphere =
πr³
We require to know the radius r to calculate the volume
using surface area of a sphere = 4πr² = 615.752
divide both sides by 4π
r² =
= 49.0248...
take the square root of both sides
r =
≈ 7, hence
V=
× π × 7³ = 1436.76 m³ → B
Answer: 0.0375
Step-by-step explanation:
0.15 divided by 4 = 0.0375
Answer:
Option B. ![x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7B9%7D%7B28%7D%28%2B%2F-%29%5Cfrac%7B%5Csqrt%7B479%7D%7D%7B28%7Di)
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
![14x^{2}+9x+10=0](https://tex.z-dn.net/?f=14x%5E%7B2%7D%2B9x%2B10%3D0)
so
substitute in the formula
Remember that
![i=\sqrt{-1}](https://tex.z-dn.net/?f=i%3D%5Csqrt%7B-1%7D)
substitute
![x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7B9%7D%7B28%7D%28%2B%2F-%29%5Cfrac%7B%5Csqrt%7B479%7D%7D%7B28%7Di)
Answer:
Step-by-step explanation:
f(- 1) = - 1 + 4 = 3
f(2) = 2 + 4 = 6
f(12) = 4(12) - 7 = 41