A. ![f(-5)=7+|-5|=12](https://tex.z-dn.net/?f=f%28-5%29%3D7%2B%7C-5%7C%3D12)
B. ![g(4)=4^3-5=59](https://tex.z-dn.net/?f=g%284%29%3D4%5E3-5%3D59)
C. ![f(0)=7+|0|=7](https://tex.z-dn.net/?f=f%280%29%3D7%2B%7C0%7C%3D7)
D. ![f(2)=7+|2|=9](https://tex.z-dn.net/?f=f%282%29%3D7%2B%7C2%7C%3D9)
E. ![g(-2)=-2^3-5=-13](https://tex.z-dn.net/?f=g%28-2%29%3D-2%5E3-5%3D-13)
Hope this helps BangtanBoyScouts
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
It is the second answer:
8x²/y -2 is a solution which is equivalent to (8x²-2y)/2. Proof:
Reduce 8x²/y -2 to the same denominator:
8x²/y - 2y/y = (8x² - 2y)/y
Depends on your state. Each state is different. I'll try to post the complete list.
Hope this is what you are looking for!