14-6+5
13
13 is the required answer
Reference to a standard normal distribution table shows that for a cumulative probability of 0.6064 the z-score is 0.27.
Answer:
He will need 38 more blueberries
Step-by-step explanation:
If each cupcake will have 6 blueberries and he wants to make 17 cupcakes he will need a total of 102 blueberries (found by multiplying 17 and 6 together). If he already has 64 he will just need to buy 38 more (found by subtracting 64 from 102). I hope this helped, if not then I apologize.
Answer:
1.A,B,C,D
2. AB, CD
3. AC, BD
4. Line AD (don't take my word for this one)
Step-by-step explanation:
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
