Y = 3/2x + 4
Add the 8 to both sides and then divide all the terms by 2
Since we want just the top 20% applicants and the data is normally distributed, we can use a z-score table to check the z-score that gives this percentage.
The z-score table usually shows the percentage for the values below a certain z-score, but since the whole distribution accounts to 100%, we can do the following.
We want a z* such that:

But, to use a value that is in a z-score table, we do the following:

So, we want a z-score that give a percentage of 80% for the value below it.
Using the z-score table or a z-score calculator, we can see that:
![\begin{gathered} P(zNow that we have the z-score cutoff, we can convert it to the score cutoff by using:[tex]z=\frac{x-\mu}{\sigma}\Longrightarrow x=z\sigma+\mu](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%28zNow%20that%20we%20have%20the%20z-score%20cutoff%2C%20we%20can%20convert%20it%20to%20the%20score%20cutoff%20by%20using%3A%5Btex%5Dz%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5CLongrightarrow%20x%3Dz%5Csigma%2B%5Cmu)
Where z is the z-score we have, μ is the mean and σ is the standard deviation, so:

so, the cutoff score is approximately 72.
This question is worded badly, so don't be alarmed.
Obviously if this were a square patter the number of red tiles could only be 1,4,9,16,25,36 etc and this is not the numbers given...It would help tremendously if they gave you another reference point so as to know how the tiles were being laid out.
If they are laid out alongside one another with a consistent border for each you wouldn't need 12 blue tiles for each red tile either...if two were laid side by side you would only need 20 blue tiles to have the border around both...you see how the arrangement makes a huge difference and you are given no details as to what that arrangement is...
So just to keep it simple, assume that each red tile is surrounded by 12 blue tiles....
Total tiles equals r+12r=13r
t=13r so
t(15)=195, t(20)=260, t(25)=325, t(30)=390
Hello!
Answer:
3(4+3g-10h)
Step-by-step explanation:
Hope this helps!