Answer:
1. I would think it would be a appropriate to use a normal model to figure this out because sometimes when you are using stuff online you can get the same number like 20 times in a row, whereas, if you use a real dice, you most likely won't roll the same number 20 times.
2. { sorry two doesn't make sense }
So uh uhm yes uhhhhhhhhhhhhhhhh
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!!!
Answer:
12.
Explanation:
6.0m/s devide it by 0.5m/s is 12.
Hope it helps. Please mark me as brainliesttt
