Answer:
P(A) = 0.2
P(B) = 0.25
P(A&B) = 0.05
P(A|B) = 0.2
P(A|B) = P(A) = 0.2
Step-by-step explanation:
P(A) is the probability that the selected student plays soccer.
Then:
P(B) is the probability that the selected student plays basketball.
Then:
P(A and B) is the probability that the selected student plays soccer and basketball:
P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:
P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.
Avoiding conviniece store purchases
317 is the answer you are looking for.
Answer:
A. Initially, there were 12 deer.
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. After 15 years, there will be 410 deer.
D. The deer population incresed by 30 specimens.
Step-by-step explanation:
The amount of deer that were initally in the reserve corresponds to the value of N when t=0
A. Initially, there were 12 deer.
B. 
B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.
C. 
C. After 15 years, there will be 410 deer.
D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:
ΔN=N(15)-N(10)
ΔN=410 deer - 380 deer
ΔN= 30 deer.
D. The deer population incresed by 30 specimens.
Answer: Option A
Step-by-step explanation:
If we have 5 cans of ketchup they cost $ 5.25. Then the unit price of cans is what each can costs individually.
The unit price should then be less than $ 5.25, and that is the price for 5 cans and we want to know the price for just one can of ketchup.
Then to calculate the price of each can divide the price of the 5 cans by 5.
