Answer:
See below
Step-by-step explanation:
When you roll an 8-sided die twice, the sample space is the set of all possible pairs (x,y) where x is the first outcome and y is the second outcome.
The sample space is:
![[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(1, 7),(1, 8)\\(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(2, 7),(2, 8)\\(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(3, 7),(3, 8)\\(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(4, 7),(4, 8)\\(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),(5, 6),(5, 7),(5, 8)\\(6, 1), (6, 2), (6, 3), (6, 4)(6, 5),(6, 6),(6, 7),(6, 8)\\(7, 1), (7, 2), (7, 3), (7, 4)(7, 5),(7, 6),(7, 7),(7, 8)\\(8, 1), (8, 2), (8, 3), (8, 4)(8, 5),(8, 6),(8, 7),(8, 8)]](https://tex.z-dn.net/?f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
The sample space of the product xy of each outcome forms the required possibility diagram.
This is given as:
Answer:
4 and then 15
Step-by-step explanation:
Answer:
135°
Step-by-step explanation:
LI = LM+MI = 35+80
Answer:
The probability is 0.3576
Step-by-step explanation:
The probability for the ball to fall into the green ball in one roll is 2/1919+2 = 2/40 = 1/20. The probability for the ball to roll into other color is, therefore, 19/20.
For 25 rolls, the probability for the ball to never fall into the green color is obteined by powering 19/20 25 times, hence it is 19/20^25 = 0.2773
To obtain the probability of the ball to fall once into the green color, we need to multiply 1/20 by 19/20 powered 24 times, and then multiply by 25 (this corresponds on the total possible positions for the green roll). The result is 1/20* (19/20)^24 *25 = 0.3649
The exercise is asking us the probability for the ball to fall into the green color at least twice. We can calculate it by substracting from 1 the probability of the complementary event: the event in which the ball falls only once or 0 times. That probability is obtained from summing the disjoint events: the probability for the ball falling once and the probability of the ball never falling. We alredy computed those probabilities.
As a result. The probability that the ball falls into the green slot at least twice is 1- 0.2773-0.3629 = 0.3576
Answer:
x 2 y -4
Step-by-step explanation:
13x = -54 -20y give it is A
-10x = 60 + 20y give it is B
A + B the sum of the left side of equation is equal to the sum of the right side of equation
13 + (-10x) = -54 -20y + 60 + 20y
3x = 60-54
3x = 6
x=2
so put the x value in A or B equation you can receive y value (B is easier)
y = -4
