Answer:
c
Step-by-step explanation:
To factor, we always have to look for common factors in the numbers given to us. In this case, in 1/2 and 6, the common factor is 1/2. Therefore, we will have to factor like so:
1/2d + 6 = 1/2 (d + 3)
Hope this helps!
The function "choose k from n", nCk, is defined as
nCk = n!/(k!*(n-k)!) . . . . . where "!" indicates the factorial
a) No position sensitivity.
The number of possibilities is the number of ways you can choose 5 players from a roster of 12.
12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
You can put 792 different teams on the floor.
b) 1 of 2 centers, 2 of 5 guards, 2 of 5 forwards.
The number of possibilities is the product of the number of ways, for each position, you can choose the required number of players from those capable of playing the position.
(2C1)*(5C2)*(5C2) = 2*10*10 = 200
You can put 200 different teams on the floor.
Answer:
In a certain Algebra 2 class of 30 students, 22 of them play basketball and 18 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?
I know how to calculate the probability of students play both basketball and baseball which is 1330 because 22+18+3=43 and 43−30 will give you the number of students plays both sports.
But how would you find the probability using the formula P(A∩B)=P(A)×p(B)?
Thank you for all of the help.
That formula only works if events A (play basketball) and B (play baseball) are independent, but they are not in this case, since out of the 18 players that play baseball, 13 play basketball, and hence P(A|B)=1318<2230=P(A) (in other words: one who plays basketball is less likely to play basketball as well in comparison to someone who does not play baseball, i.e. playing baseball and playing basketball are negatively (or inversely) correlated)
So: the two events are not independent, and so that formula doesn't work.
Fortunately, a formula that does work (always!) is:
P(A∪B)=P(A)+P(B)−P(A∩B)
Hence:
P(A∩B)=P(A)+P(B)−P(A∪B)=2230+1830−2730=1330
