Answer:
0.25
Step-by-step explanation:
We have a total of ten student, and three students are randomly selected (without replacement) to participate in a survey. So, the total number of subsets of size 3 is given by 10C3=120.
On the other hand A=Exactly 1 of the three selected is a freshman. We have that three students are freshman in the classroom, we can form 3C1 different subsets of size 1 with the three freshman; besides B=Exactly 2 of the three selected are juniors, and five are juniors in the classroom. We can form 5C2 different subsets of size 2 with the five juniors. By the multiplication rule the number of different subsets of size 3 with exactly 1 freshman and 2 juniors is given by
(3C1)(5C2)=(3)(10)=30 and
Pr(A∩B)=30/120=0.25
1/2 or 6/12. I think at least
Step-by-step explanation:
YO ≅ NZ
Given
YO + OZ ≅ NZ + OZ
Additive property
YZ ≅ NO
Segment addition postulate
∠M ≅ ∠X, ∠N ≅ ∠Y
Given
ΔMNO ≅ ΔXYZ
AAS congruence
Answer: it's 6x + 8 if you solve it
Answer:
There are three terms in the simplified expression.
Step-by-step explanation:
We have to simplify the expression and have to count the number of terms that the expression has.
The expression is 4y - 6 + y² - 9
= 4y + y² - 6 - 9
= y² + 4y - 15
Therefore, there are three terms in the simplified expression, one for y² term, another is y term and the constant term. (Answer)