Answer:
19/54
Step-by-step explanation:
- 7/27 + 11/18
LCM of the 27, 18 is 54
- 7/27 * (2/2) + 11/18 * (3/3)
- 14/54 + 33/54
33 - 14/54
19/54
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Answer:
sod- 1662.6
fencing- 807.6
all- 2470.2
Step-by-step explanation:
sod-
25.5 * 8.25 = 207.825
207.825*8= 1662.6
fencing-
8.25+8.25+25.5+25.5= 67.3
67.3*12= 807.6
fencing + sod
1662.6 + 807.6= 2470.2
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:
The interquartile range is from Q1 to Q3 and to get this you have to subtract Q2 by Q3. The 10 units represent that only 10 units are fit in the given range.
Step-by-step explanation:
given,velocity = 5 m/s
and distance=12×1000=12000m
now,time =?
we have , v=s/t
or, t= s/v
so,t=12000/5
2400sec.....ans
=