250 students complete a business college diploma. The results are normally distributed with a mean of 64% and a standard deviation of 8.6%. 15 students gain distinctions, 35 gain credits and 23 fail the course. Determine the cut-off percentages for each of the grades: a) Pass b) Credit c) Distinction

fail 0-x. mid values. =X/2

pass x-a. =(x+a)/2. pass no.=250-35-15-23=177

credit a-y. =(a+y)/2

distinction y-100. =(y+100)/2

mean=1/250*(x/2)×23+(x+a)/2*177+(y+a)/2×35+(y+100)/2×15=64

1/500[23x+177x+177a+35a+35y+15y+100*15)=64

200x+212a+50y+1500=32000

variance=[(X/2-64)^2*23+(x+a,/2-64)^2*277+(a+y/2-64)^2*35+(100+y/2-64)^2*15]/250=(8.6)^2

I've made two equations that has to be simplified.

probably a nother data has to be given since we have 3 variables to solve for(X,a,y)

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this