Y= a(x-3)^2+6
2= a(0-3)^2+6
2=a(-3)^2+6
2=a(9)+6
2-6=9a
-4=9a
-4/9=a
Therefore the equation in vertex form is
y = -4/9 (x-3)^2+6
Answer:
4/8
Step-by-step explanation:
8 is how many cookies in total and 4 is how many that needs to be distributed.
Answer:
<h2>2/5</h2>
Step-by-step explanation:
The question is not correctly outlined, here is the correct question
<em>"Suppose that a certain college class contains 35 students. of these, 17 are juniors, 20 are mathematics majors, and 12 are neither. a student is selected at random from the class. (a) what is the probability that the student is both a junior and a mathematics majors?"</em>
Given data
Total students in class= 35 students
Suppose M is the set of juniors and N is the set of mathematics majors. There are 35 students in all, but 12 of them don't belong to either set, so
|M ∪ N|= 35-12= 23
|M∩N|= |M|+N- |MUN|= 17+20-23
=37-23=14
So the probability that a random student is both a junior and social science major is
=P(M∩N)= 14/35
=2/5
I would say Jupiter but i may be wrong
I saw the images that accompanied this problem.
2 circles: circle 1 has 6 slices ; circle 2 has 3 slices
Circle 1 is for Tom: you can shade 4 slices out of the 6 slices.
Circle 2 is for Liz: you can shade 2 slices out of the 3 slices.
4/6 is equal to 2/3 because when 4 and 6 are both divided by 2, they give quotient of 2 and 3, respectively.
4 ÷ 2 = 2
6 ÷ 2 = 3
