It's a probability problem to find the odds of picking a green or red
shirt out of the 10 shirts on Thursday, Friday and Saturday since you
have randomly already know you have picked a blue shirt on the other
days. Not sure if you have this as a multiple question problem as you didn't list any possible answers (A. 7/20, B. 5/47, C. 2/5, D. 4/125) to the question. A, B, C, D being like 7 chances out of 20, 5 chances out 47, 2 chances out of chances 5 or 4 chances out of 125 (example answers only).
Probability = Number favorable outcomes / total number of outcomes
Answer:
64 different black jack hands
Step-by-step explanation:
there are 4 aces in a deck, and you can score a blackjack if any of these 4 aces is paired with any of the 16 different cards that are worth 10 points. Total number of different blackjack combinations = 4 aces x 16 different tens = 64.
Even though the number of possible blackjack hands might seem large, the possibilities of actually getting one are really low:
possibility of getting 1 ace = 4/52 = 1/13
possibility of getting 1 ten = 16/51
possibility of getting a blackjack = 16 / 663
if you start with the 10 first the odds are the same:
possibility of getting 1 ten = 16/52 = 4/13
possibility of getting 1 ace = 4/51
possibility of getting a blackjack = 16 / 663
Answer:
The number of child tickets was 94
Step-by-step explanation:
Let
x-----> the number of child tickets
y----> the number of adult tickets
we know that
-----> equation A
-----> equation B
Solving by graphing
The solution of the system of equations is the intersection point both graphs
The intersection point is (94,57)
see the attached figure
therefore
The number of child tickets was 94
The number of adult tickets was 57
the answer is eighteen (you should know that)
