There are 100 students each enrolled in at least one of three science classes. of those students, 60 are enrolled in chemistry, 45 in physics, and 30 in biology. some students are enrolled in two science classes, and 10 students are enrolled in all three. (a) how many students are enrolled in exactly two science classes? (b) there are 9 students taking both chemistry and physics (but not biology), and 4 students taking both physics and biology (but not chemistry). how many are taking both chemistry and biology (but not physics)? lewis, harry. essential discrete mathematics for computer science (p. 57). princeton university press. kindle edition.

  a) 15

  b) 2

Step-by-step explanation:

a) The sum of the enrollments in chemistry (60), physics (45), and biology (30) counts those triply enrolled 3 times and those doubly-enrolled twice. This sum will exceed the total number of students by 1 times those double-enrolled and twice those triply-enrolled.

We know that there are 10 students triply-enrolled, so the difference ...

  (60 +45 +30) -2(10) = 15

is the number of doubly-enrolled students.

There are 15 students enrolled in exactly 2 science classes.

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b) There are 9+4 = 13 students doubly-enrolled in physics and something else. Using the result from part A, there will be 15 -13 = 2 students doubly-enrolled in chemistry and biology, but not physics.

a) 15

b) 2

Step-by-step explanation:

There are 15 students in 2 science classes.

2 students enrolled in both chemistry and biology, but not physics.

15 -13 = 2

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

b.-4 units

all you   do is count how many times you move