Answer:
0.281 = 28.1% probability a given player averaged less than 190.
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.
A bowling leagues mean score is 197 with a standard deviation of 12.
This means that ![\mu = 197, \sigma = 12](https://tex.z-dn.net/?f=%5Cmu%20%3D%20197%2C%20%5Csigma%20%3D%2012)
What is the probability a given player averaged less than 190?
This is the p-value of Z when X = 190.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{190 - 197}{12}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B190%20-%20197%7D%7B12%7D)
![Z = -0.58](https://tex.z-dn.net/?f=Z%20%3D%20-0.58)
has a p-value of 0.281.
0.281 = 28.1% probability a given player averaged less than 190.
The correct answer is A. x=22
Since the angles of a rectangle are right angles (90 degrees) you would set it up as 90=5x-20. Then combine like terms and solve the equation. (If you need me to show you how step by step tell me)
Answer:
im not so positive sorry
Step-by-step explanation:
Answer:
ZERO
Step-by-step explanation
Write the given function as
F(x) = ax + b
i.e Y = ax + b
now differentiate Y with respect to X
dY/dX = d (ax + b)/ dx
dY/dX = a
now again differentiate this function with respect to x
so, d^2Y/dX^2 = d^2 (a)/dX^2
= 0 (ZERO)
note: differentiation of constant function is zero
Can you please take a picture of each pair of points