Using the normal distribution, it is found that 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
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 above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation for the amounts are given as follows:
![\mu = 3456, \sigma = 478](https://tex.z-dn.net/?f=%5Cmu%20%3D%203456%2C%20%5Csigma%20%3D%20478)
The proportion is the <u>p-value of Z when X = 4250</u>, hence:
![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{4250 - 3456}{478}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B4250%20-%203456%7D%7B478%7D)
Z = 1.66
Z = 1.66 has a p-value of 0.9515.
Hence 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
More can be learned about the normal distribution at brainly.com/question/15181104
#SPJ1
Answer:
![\sqrt{8x-4}](https://tex.z-dn.net/?f=%5Csqrt%7B8x-4%7D)
Step-by-step explanation:
Substitute x = g(x) into f(x) , that is
f(g(x))
= f(8x - 13)
= ![\sqrt{8x-13+9}](https://tex.z-dn.net/?f=%5Csqrt%7B8x-13%2B9%7D)
= ![\sqrt{8x-4}](https://tex.z-dn.net/?f=%5Csqrt%7B8x-4%7D)
Answers:
1.) 78 would be the range
3.) 18.8 or letter C would be correct
4.) 9.4
Answer:
OB) 2:1
Step-by-step explanation:
5 can go into 10 twice and 5 can go into itself once