Answer:
This might not be correct though. I'm only in middle school lol. I am sure about the last one.
Well, 1/3 is equal to 0.3, and 1/4 is equal to .25, so it can't be a.
3/4 is equal to .75 and 6/9 is equal to 0.6, so it can't be b.
5/8 is equal to 0.625 and 20/32 is also equal to 0.625, so it could be c.
7/9 is equal to 0.7 and 21/28 is equal to 0.75, so it can't be D.
Your best bet is C. Because both sides of C are equal.
C 5/8=20/32
Answer:
Part A
W W W M W W T W W L W W
W W M M W M T W M L W M
W W T M W T T W T L W T
W W L M W L T W L L W L
W M W M M W T M W L M W
W M M M M M T M M L M M
W M T M M T T M T L M T
W M L M M L T M L L M L
W T W M T W T T W L T W
W T M M T M T T M L T M
W T T M T T T T T L T T
W T L M T L T T L L T L
W L W M L W T L W L L W
W L M M L M T L M L L M
W L T M L T T L T L L T
W L L M L L T L L L L L
Part B
There are 64 possible outcomes. The sample size is 64.
Part C
To find the probability that Erin drinks lemonade one day, tea one day, and water one day, consider all the cases in which L, T, and W occur one time. Because the order doesn't matter in this scenario, these six outcomes from the list represent the desired event: W T L, T W L, T L W, W L T, L W T, and L T W.
The size of the sample space is 64. So, the probability that Erin drinks lemonade one day, tea one day, and water one day is 3/32.
Part D
To find the probability that Erin drinks water on two days and lemonade one day, we consider all the cases in which two Ws and one L occur. Because the order doesn't matter in this scenario, these three outcomes from the list represent the event: W W L, W L W, and L W W.
The size of the sample space is 64. So, the probability that Erin drinks water two days and lemonade one day is 3/64
Step-by-step explanation:
