Of the students in the college, 60% of the students reside in the hostel and 40% of the students are day scholars. Previous year result reports that 30% of all students who stay in the hostel scored A Grade and 20% of day scholars scored A grade. At the end of the year, one student is chosen at random and found that he/she has an A grade. What is the probability that the student is a hostlier?

9 / 13

Step-by-step explanation:

From the question :

Let :

Hostel students = h

Day students = d

Students with A grade = a

P(h) = 60% = 0.6

P(d) = 40% = 0.4

P(hostel with A grade) ; P(A / H) = 0.3

P(day with A grade) ; P(A / D) = 0.2

The probability that a student picked at random withbA grade is an hostel student is given as :

Probability of being an hostel student gun he has A :

P(H | A) = (P(A/H) * P(H)) ÷ P(A/H) * P(H) + P(A/D) * P(D))

P(H | A) = (0.3 * 0.6) ÷ ((0.3*0.6) + (0.2*0.4))

P(H | A) = 0.18 ÷ (0.18 + 0.08)

P(H | A) = 0.18 ÷ 0.26

P(H | A) = 9 / 13

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.