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
I am assuming you're doing a Punnet square, so you multiply each one. So on the top left box, it is -45x^2, the top right is 9x, bottom left is 35x, and bottom right is -7.
I wrote some notes please read and if you not sure contact me
Percentages are parts of 100, so to convert 0.8% to a decimal, we just divide 0.8 by 100.
0.8 / 100 = 0.008
Dividing by 100 is the same as moving the decimal place two times to the left.
Answer:
68 degrees
Step-by-step explanation:
90 - 22 = 68 degrees
when you see the square looking thing it indicated a right angle which is equal to 90 degrees :)
