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.
Answer:
bro just did this on a TGA
Step-by-step explanation:
A. no solutions and a contradiction
B. infinate solutions and an identity
C. one solution and niether
Answer:
There are no real solutions to this equation because the square root of a negative number is not real. So answer B
Step-by-step explanation:
<u>x / 5 = y</u>
Multiply each side by 5 : x = 5y
Multiply each side by 2 : <em>2x = 10y</em>
The expression 4x^2represents 4.x.x