The event "Atleast once" is the complement of event "None".
So, the probability that Marvin teleports atleast once per day will the compliment of probability that he does not teleports during the day. Therefore, first we need to find the probability that Marvin does not teleports during the day.
At Morning, the probability that Marvin does not teleport = 2/3
Likewise, the probability tha Marvin does not teleport during evening is also 2/3.
Since the two events are independent i.e. his choice during morning is not affecting his choice during the evening, the probability that he does not teleports during the day will be the product of both individual probabilities.
So, the probability that Marvin does not teleport during the day = 
Probability that Marvin teleports atleast once during the day = 1 - Probability that Marvin does not teleport during the day.
Probability that Marvin teleports atleast once during the day = 
Consider point P(x,y) such that P, X and Y are collinear,
As vectors
XP = XO + OZ where O(0,0)
XP = OZ - OX
XP= (x,y) - (-3,3)
XP = (x+3, y-3)
Similarly,
PY = (6-x, -3-y)
But XP= 2^PY
[x+3, y-3] = [2(6-x), 2(-3-y)]
Given both vectors are equal, as they go in the same direction, Solve for x and y accordingly:
x+3 = 12 - 2x
x = 3
y-3 = -6-2y
y = -1
Therefore, P(3,-1)
The correct answer is 3.53
I can not set up a 2 way table online but what they want is a collom of the information neatly laid out...
With the 9th graders in one collom and the 10th graders in the other
like this
9th grader 10th grader
# of students 32 40
# of participated 32 18
Draw this out with lines and it will be correct.
I hope this helps GOOD LUCK...this is tough math for a middle schooler.
Answer:
3/5
Step-by-step explanation:
60 percent is the fraction 60/100 and simplifyping that by ten is 6/10
6/10 divided by two is 3/5