Number of tickets: T.
Number of customers: c
Initially the number of tickets is T0=150, when the group hasn't sold any tickets (c=0). Then the graph must begin with c=0 and T=150. Point=(0,150). Possible options: Graph above to the right and graph below to the left.
They sell the tickets in pack of three tickets per customer c, then each time they sell a pack of three tickets to a customer, the number of tickets is reduced by 3 (-3c). Then the number of tickets, T, the group has left after selling tickets to c customers is:
T=150-3c→T=-3c+150
For T=0→-3c+150=0→150=3c→150/3=c→c=50. The graph must finish with c=50, T=0. Final point=(c,T)=(50,0)
Answer:
The correct graph is above to the right, beginning on vertical axis with T=150 and finishing on horizontal axis with c=50.
The correct equation is T=-3c+150
3/4 = or .75
Divide the numerator and denominator to reduce the fraction.
Probability that a student will play both is 7/30
Step-by-step explanation:
Total students = 30
No. of students who play basketball = 18
Probability that a student will play basketball = 18/30
= 3/5
No. of students who play baseball = 9
Probability that a student will play baseball = 9/30
= 3/10
No. of students who play neither sport = 10
Probability that a student will play neither sport = 10/30
= 1/3
To find :
Probability that a student will play both = p(student will play both)
No.of students who play sport = 30 - 10
= 20
Out of 20 students 18 play basketball and 9 play baseball.
So, some students play both the sports.
No. of students who play both sports = 18 + 9 - 20
= 7
p(student will play both) = 7/30
Probability that a student will play both is 7/30
Answer:
Part A: hx2=a
Part B: 5hours
Step-by-step explanation:
Answer:
6 is the answer
Step-by-step explanation:
Hope I have helped.
