That's a 30-60-90 triangle, so the hypotenuse will always be 2x, the shorter leg will be x and the longer leg will be x√3. In this case, we are given the longer leg and we can deduce that x = 9. So, a = x = 9 and c = 2x = 18. Your answer is A, a = 9 yd, c = 18 yd.
There are 45 numbers between 49 and 95
Numbers in the fifties: 50,51,52,53,54,55,56,57,58,59
Numbers in the sixties: 60,61,62,63,64,65,66,67,68,69
Numbers in their seventies: 70,71,72,73,74,75,76,77,78,79
Numbers in their eighties: 80,81,82,83,84,85,86,87,88,89
Numbers in their nineties: 90,91,92,93,94
Multiples of 2:
50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94
Multiples of 3:
51,54,57,60,63,66,69,72,75,78,81,84,87,90
Multiples of 9:
54,63,72,81,90
Notice that all multiples of nine are also multiples of three? That is because three is a factor of nine
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:
