If the points are on the same line they are co linear. when working with coordinates if the all have the same y as in (x,y) they are on the same line for example, (4,5) (2,5) (9,6) are collinear
Answer:
a.) 0.7063
b.) 23
Step-by-step explanation:
a.)
Let X be an event in which at least 2 students have same birthday
Y be an event in which no student have same birthday.
Now,
P(X) + P(Y) = 1
⇒P(X) = 1 - P(Y)
as we know that,
Probability of no one has birthday on same day = P(Y)
⇒P(Y) =
where there are n people in a group
As given,
n = 30
⇒P(Y) =
= 0.2937
∴ we get
P(X) = 1 - 0.2937 = 0.7063
So,
The probability that at least two of them have their birthdays on the same day = 0.7063
b.)
Given, P(X) > 0.5
As
P(X) + P(Y) = 1
⇒P(Y) ≤ 0.5
As
P(Y) =
We use hit and trial method
If n = 1 , then
P(Y) =
= 1
0.5
If n = 5 , then
P(Y) =
= 0.97
0.5
If n = 10 , then
P(Y) =
= 0.88
0.5
If n = 15 , then
P(Y) =
= 0.75
0.5
If n = 20 , then
P(Y) =
= 0.588
0.5
If n = 22 , then
P(Y) =
= 0.52
0.5
If n = 23 , then
P(Y) =
= 0.49
0.5
∴ we get
Number of students should be in class in order to have this probability above 0.5 = 23
5w + 8 - w = 6w - 2(w -4).
Let's reduce the equation on the left:
4w + 8; (1)
and now let's reduce the equation on the right
6w - 2w +8 = 4w +8 (2).
We notice that (1) ≈ (2), and this is the only IDENTITY
A quadrilateral MUST be a parallelogram if it has one pair of opposite sides both parallel and congruent. A quadrilateral MUST be a parallelogram if it has both pairs of its opposite angles congruent (or equal in measure). A quadrilateral MUST be a parallelogram if it has both diagonals bisecting each other.
Answer:
Hi... Your answer is... B