Answer:
2
Step-by-step explanation:
We consider the x- and y-coordinates separately. Let the coordinates of G be (x, y). Now considering the x-coordinates:
FG/FH = (x - (-3)) / (-3 - (-3)) = 2/3
x + 3 = (2/3)(6)
x = 1
For the y-coordinates:
FG/FH = (y - 2) / (7 - 2) = 2/3
y - 2 = (2/3)(5)
y = 16/3
Therefore the coordinates of G are (1, 16/3).
He could only make one team, considering that if you do 12-9, you are left with 3, and Jason wants to have his team consisted of 9 players. However, if you make 3 teams with 4 players on each team, or 4 teams with 3 players on each team, you could do that as well.
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
5x-50= 30-15x
20x= 80
x= 4 is the solution to this problem
