All tutors that are employed at a local college must take a test and must score in the top 20%. if the scores are normally distributed with a mean of 78 and a standard deviation of 5, what is the cutoff score to be hired?

To solve this problem, we use the z statistic with formula:

z = (x – u) / s

or

x = s z + u

First we find the value of z at P = top 20%

z = 0.84

So the cut-off is:

x = 5 * 0.84 + 78

x = 82.2                 (ANSWER)

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

exactly you go it

step-by-step explanation: