Question

150 college freshmen were interviewed
85 were registered for a Math class
70 were registered for an English class
50 were registered for both Math and English
a) How many signed up only for a Math Class?
b) How many signed up only for an English Class?
c) How many signed up for Math or English?
d) How many signed up neither for Math nor English?

Answer 1

Step-by-step explanation:

85-50 =35... {as 50 is intersection set} so Onlyy fr maths, the no. Is 35.

Similarly, 70-50= 20

Or is union set acc to my knowledge so..

Let n(E) be fr eng

And n(M)fr math

N(EUM) =n(E) + n(M) - n(E intersection M)

150= 85+70-50

150= 155-50

150= 105.

150-105= x

X=45.

Hence none of them signed = 150- 45- 35- 20

= 150- 80- 20

= 150-100

= 50.

Ty.

Answer 2

Answer:

a) 35

b) 20

c) 85 or 75

d) 45

Step-by-step explanation:

a) Let only math = x and math = a and english = b and math english = ab

   x = a - ab

   x = 85 - 50

   35

b) Let only english = x and math = a and english = b and math english = ab

   x = b - ab

   x = 70 - 50

   20

c) 85 or 75

d) Let the not registered in math and english = x and only math = a and only english = b and math english = ab and total of student that interviewed = c

   x = c - a + b + ab

   x = 150 - 35 + 20 + 50

   x = 150 - 105

   x = 45