<span>The 3rd astronaut would catch the 2nd astronaut and throw the 2nd astronaut towards the 1st and the game would end there.
The key thing to remember is conservation of momentum. Since all of the astronauts have the same mass and strength, I will be introducing a unit called "A" which represents the maximum momentum that one astronaut can produce while throwing another. So here's the game of catch, throw by throw.
Before the game begins, I will assume all three astronauts are stationary and have 0 momentum. So
Astronaut 1 = 0 A (Stationary, next to astronaut 2)
Astronaut 2 = 0 A (Stationary, next to astronaut 1)
Astronaut 3 = 0 A (Stationary)
1st astronaut grabs the 2nd astronaut and throws him towards the 3rd.
Since every action has an equal and opposite reaction, what will happen is the 1st astronaut will be sent moving backwards with a momentum of -1/2A and the 2nd astronaut will be heading towards the 3rd with a momentum of +1/A. So we're left with
Astronaut 1 = -1/2 A (Moving to the left)
Astronaut 2 = +1/2 A (Moving to the right)
Astronaut 3 = 0 A (Stationary)
Now the 3rd astronaut catches the 2nd who was thrown at him. Both of them continue moving in the same direction as the 2nd astronaut was just prior to being caught, but at a reduced velocity, giving
Astronaut 1 = -1/2 A (Moving to the left)
Astronaut 2 = +1/4 A (Moving to the right, slowly)
Astronaut 3 = +1/4 A (Moving to the right, slowly)
Finally, Astronaut 3 throws astronaut 2 back towards Astronaut 1, giving
Astronaut 1 = -1/2 A (Moving to the left)
Astronaut 2 = +1/4 A -1/2A = -1/4A (Moving to the left, slowly)
Astronaut 3 = +1/4 A +1/2A = +3/4A (Moving to the right, rapidly)
So what you're left with is Astronaut 1 moving to the left faster than Astronaut 2, so those two astronauts will never catch each other. Meanwhile, Astronaut 3 is moving to the right and getting further and further away from the other 2 astronauts. So none of the astronauts will ever be able to catch or throw anyone ever again.</span>
Answer: Our required probability would be 0.70.
Step-by-step explanation:
Since we have given that
Number of players = 14
Number of players have recently taken a performance enhancing drug = 3
Number of players have not recently taken a performance enhancing drug = 14-3=11
Number of members chosen randomly = 5
We need to find the probability that at least one of the tested players is found to have taken a performance enhancing drug.
P(Atleast 1) = 1 - P(none is found to have taken a performance enhancing drug)
So, P(X≥1)=1-P(X=0)
Hence, our required probability would be 0.70.
1/18k^6 so uhhhh ya laalalallaa
For 5, what is happening is all the points shift down negative 2 on the Y axis and then flip over the x axis.
For 6, it would flip over the Y axis and start at positive 1 x the opposite
It has to do with the unit circle, I think it is A because it should be 2(0+ (-1)) but I would probably wait for someone to come behind me and explain it better sorry couldn't help more
