5/6 as a percent is 83.33%
Set x as adult tickets.
Set y as children's tickets.
x + y = 15
30x + 20y = 270
Solve for x in the first equation.
x + y = 15
x = 15 - y
Plug this into the second equation.
30x + 20y = 270
30(15 - y) + 20y = 270
450 - 30y + 20y = 270
450 - 10y = 270
-10y = -180
y = 18
If there is 18 childrens tickets, there should be -3 adult tickets.
This is impossible, and this impossible answer occured because the question is written wrong.
There are a total of 15 tickets
The smallest costing ticket is the childrens ticket, which costs 20$.
If he only bought children tickets, this would be 20x15 which is 300$.
300$ is over 270$, which makes the question impossible.
Answer:
0.64
Step-by-step explanation:
P(J / R) = P (J and R) / P(R)
0.8 = P (J and R) / 0.6
P (J and R) = 0.6 * 0.8 = 0.48 [Probability John practicing and it is raining]
P(J / NR) = P (J and NR) / P(NR)
0.4 = P (J and NR) / (1 - 0.6) = P (J and NR) / 0.4
P (J and NR) = 0.4 * 0.4 = 0.16 [Probability John practicing and it is not raining]
Hence;
Probability of John practicing regardless of weather condition is
P(John Practicing) = 0.48 + 0.16 = 0.64
HOPE THIS HELPED!!!
