Answer:
The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
Step-by-step explanation:
Problems of normally distributed samples are solved using 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 problem, we have that:
![\mu = 125, \sigma = 24](https://tex.z-dn.net/?f=%5Cmu%20%3D%20125%2C%20%5Csigma%20%3D%2024)
What is the reading speed of a sixth-grader whose reading speed is at the 90th percentile
This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.
![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 - 125}{24}](https://tex.z-dn.net/?f=1.28%20%3D%20%5Cfrac%7BX%20-%20125%7D%7B24%7D)
![X - 125 = 1.28*24](https://tex.z-dn.net/?f=X%20-%20125%20%3D%201.28%2A24)
![X = 155.72](https://tex.z-dn.net/?f=X%20%3D%20155.72)
The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
Answer:
The cost of one pass is 25$
Step-by-step explanation:
Call the cost of one of the passes, P. So we have
63 = 13 + 2P - subtract 13 from both sides
50 = 2P - divide both sides by 2
25 = P
So the cost of one pass = $25
Just watch videos on youtube.
Answer:
x=7
Step-by-step explanation:
4(x-8)= -2(x-5)
4x+32= -2x+10
+2x +2x
6x-32=10
+32 +32
6x= 42
divide by 6 on both sides
x=7