In a certain college class, 40% of the admitted students were in top 10% of their high school class, 25% were in the next 10%, and the remaining 35% were below the top 20%. of these students, 90%, 70%, and 20% were passing this course, respectively. if a randomly selected student is failing, then what is the probability that this student was below 20% of his or her high school class (round off to second decimal place)?

  0.71

Step-by-step explanation:

The fraction of students failing is the sum of the fractions of students failing in each category. Those fractions are ...

  top 10%: 0.40 × (1 -0.90) = 0.04

  next 10%: 0.25 × (1 -0.70) = .075

  bottom 80%: 0.35 × (1 -0.20) = 0.28

So, the total fraction of students failing is ...

  0.04 +0.075 +0.28 = 0.395

The desired probability is ...

  p(bottom 80% | failing) = p(bottom 80% & failing) / p(failing)

  = 0.28/0.395 ≈ 0.7089 ≈ 0.71

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204