Answer:
Sorry it's acually (h+3)(h-9)
Answer and explanation:
The gambler's fallacy is the fallacy of belief that if an event such as a loss occurs more frequently in the past, it is less likely to happen in the future. We assume here that this belief is true, therefore
If she loses, her probability of winning increases =3/4
If she wins, her probability to win is normal =1/2
Given that probability of winning is 1/2
Probability of losing is 1-1/2=1/2
Probability that she wins the tournament is probability that she wins the first two games and loses the last or wins the first game, loses the second and wins the last or loses the first game and wins the last two games or probability that she wins all three games
=1/2*1/2*1/2+1/2*1/2*3/4+1/2*3/4*1/2+1/2*1/2*1/2
=25/48
Probability of winning the tournament if she loses the first game
=1/2*3/4*1/2= 3/16
Note: whenever there is "or" in probability, you add
Hello!
The graph y < -2/3x + 2 has a dotted line. That means that any points on the dotted line or not shaded, is not a solution to this inequality. Since we are given 4 choices, we can substitute those values into the given inequality to see if it is true.
A) (0, 2)
2 < -2/3(0) + 2
2 < 2, this is false.
B) (3, 0)
0 < -2/3(3) + 2
0 < -2 + 2
0 < 0, this is false.
C) (0, 0)
0 < -2/3(0) + 2
0 < 2, this is true.
D). (0, 3)
3 < -2/3(0) + 2
3 < 2, this is false.
To check if choice C) (0, 0) is true, we should look at the given graph.
Since (0, 0) is in the shaded area, and is not graphed on the dotted line, therefore, a solution to this linear inequality is C, (0, 0).
Answer:
The 3rd option is correct.
Step-by-step explanation:
This is because of an exponent rule that goes:
(xy)^z = x^z(y^z)
The whole thing gets raised to the power. Therefore only the 3rd one works (ps. if you can mark as brainliest that would be amazing, but if not that's fine)
