<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:
33.3% probability that both children are girls, if we know that the family has at least one daughter named Ann.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
The family has two children.
The sample space, that is, the genders of the children may be divided in the following way, in which b means boy and g means girl.
b - b
b - g
g - b
g - g
We know that they have at least one girl. So the sample space is:
b - g
g - b
g - g
What is the probability that both children are girls, if we know that the family has at least one daughter named Ann?
Desired outcomes:
Both children being girls, so
g - g
1 desired outcome
Total outcomes
b - g
g - b
g - g
3 total outcomes
Probability
1/3 = 0.333
33.3% probability that both children are girls, if we know that the family has at least one daughter named Ann.
Answer:
163.2
Step-by-step explanation:
D
Answer:
go to this link
Step-by-step explanation:
Videos
2:41
Manipulating expressions using structure (example 2) (video)
Answer:
C. 55
Step-by-step explanation:
Since all angles in a triangle add up to 180°, we have:
40 + y + y + 30 = 180
2y + 70 = 180
2y = 110
y = 55
