In the question, it is already given that the total number of runners in the race is 60 and out of them 1/3 dropped out in the first half. In the second half 1/4 of the remaining runners dropped out.
Now
Total number of runners in the race = 60
Number of runners that dropped out in the first half = 1/3 * 60
= 20
Number of runners remaining = 60 - 20
= 40
Number of runners dropping out in the second half = 40 * 1/4
= 10
Then the number of runners that finished the race = 40 - 10
= 30
Then 30 runners completed the race.
Answer:
No
Step-by-step explanation:
If you add all the sales you do not meet your goal of $25,000
Answer:
a) P ( E | F ) = 0.54545
b) P ( E | F' ) = 0
Step-by-step explanation:
Given:
- 4 Coins are tossed
- Event E exactly 2 coins shows tail
- Event F at-least two coins show tail
Find:
- Find P ( E | F )
- Find P ( E | F prime )
Solution:
- The probability of head H and tail T = 0.5, and all events are independent
So,
P ( Exactly 2 T ) = ( TTHH ) + ( THHT ) + ( THTH ) + ( HTTH ) + ( HHTT) + ( HTHT) = 6*(1/2)^4 = 0.375
P ( At-least 2 T ) = P ( Exactly 2 T ) + P ( Exactly 3 T ) + P ( Exactly 4 T) = 0.375 + ( HTTT) + (THTT) + (TTHT) + (TTTH) + ( TTTT)
= 0.375 + 5*(1/2)^4 = 0.375 + 0.3125 = 0.6875
- The probabilities for each events are:
P ( E ) = 0.375
P ( F ) = 0.6875
- The Probability to get exactly two tails given that at-least 2 tails were achieved:
P ( E | F ) = P ( E & F ) / P ( F )
P ( E | F ) = 0.375 / 0.6875
P ( E | F ) = 0.54545
- The Probability to get exactly two tails given that less than 2 tails were achieved:
P ( E | F' ) = P ( E & F' ) / P ( F )
P ( E | F' ) = 0 / 0.6875
P ( E | F' ) = 0
5 you add them all up and then divide by how many number you added
(6+3+2+1+9+9) and then divide by 6
I’m not sure for y=x but for when x=6 he slope is undefined and there is no y-intercept
