Answer: B. 31
Step-by-step explanation:
This linear regression was constructed by relating the hours practiced per week and the number of competitions won.
Going by this graph, the number of competitions they can expect to win at 5 practices a week is 31.
This is derived by looking for the point where 5 competitions on the x-axis intersects with the line. This point is at 31 competitions on the y axis which would make it the answer.
Let's assume
number of child tickets sold on saturday is x
number of adult tickets sold on saturday is y
On saturday, twice as many adult tickets as child tickets were sold
so, we get

child admission is $6.30 and adult admission is $9.60
a total sales of $535.50
we get

now, we can plug y=2x

now, we can solve for x


now, we can find y


so,
number of child tickets sold on saturday is 21
number of adult tickets sold on saturday is 42.........Answer
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
Look at the solution ( again me)