Answer:
answer is a but please mark me brainliest if I m correct
The expected value of the first game is -$0.50 and of the second game is -$0.52.
There are 10³ possible numbers for the lottery, and only 1 of them will match in the correct order; this gives a probability of 1/1000. To find the expected value, we multiply this by the winnings (499 after the $1 cost); we also multiply the probability of losing (999/1000) by the amount lost (-1):
1/1000(499)+999/1000(-1)
499/1000 - 999/1000 = -500/1000 = -0.50
For the second game, since the number is "boxed", there are 3! ways to get the correct digits; this gives a probability of 6/1000. Multiply this by the winnings, 79 (after the $1 cost); multiply the probability of losing (994/1000) by the loss (-1):
6/1000(79) + 994/1000(-1) = 474/1000 - 994/1000 = -520/1000= -0.52
X= 3
4*9 -18 =18
3*9-9=18
you had to set the equation equal to each other
Step-by-step explanation:
no
no
yes
yes
hope this helps
Step-by-step explanation:
sinΘ=-3,cosΘ=4,cotΘ=-4/3,secΘ=1/4 & cosecΘ=-1/3
