If AB bisects 2 CAF, and angle EAF = 66°, then the angle BAF =
1 answer:
You might be interested in
Answer:
The probability is 0.0052
Step-by-step explanation:
Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:
P(A/B) = P(A∩B)/P(B)
The probability P(B) that at least three are aces is the sum of the following probabilities:
- The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
- There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:
So, the number of ways to select exactly 3 aces is:
Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725
Then, the probability P(B) that at least three are aces is:
On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:
P(A∩B) = 1/270,725
Finally, the probability P(A/B) that all four are aces given that at least three are aces is:
Answer:
Option (2). 1
Step-by-step explanation:
Coordinates of point A, B, C and D are,
A(-4, 4), B(-2, 4), C(-2, 1) and D(-4, 3).
Quadrilateral ABCD when rotated 90° clockwise about the origin,
Rule for the rotation of the vertices,
(x, y) → (y, -x)
Following the rule of rotation coordinates of the image points,
A(-4, 4) → A'(4, 4)
B(-2, 4) → B'(4, 2)
C(-2, 1) → C'(1, 2)
D(-4, 3) → D'(3, 4)
Since all image points have the positive coordinates (x and y coordinates), image quadrilateral A'B'C'D' will be located in 1st quadrant.
Option (2) is the correct option.
