The numbers between 76 and 88 are: {77,78,79,80,81,82,83,84,85,86,87}.
0-1/4-2/3-3/4-4/4
0-1/8/2/8-3/8-4/8-5/8-6/8-7/8
Answer:
about 9.82
Step-by-step explanation:
i just divided 432 by 44 in order to isolate x
i hope that correct ans helps
I'll walk you through the first one, and then you should be able to do the rest.
The first step is to right out your numbers from least to greatest
Problem #1: 2,4,5,6,8,10,13,17,19,20
To find the MEDIAN, you're simply crossing out a number from each end until you meet in the middle. In this case, you have an even number of data, which means once you get to the middle, you're have to find the average.
In this problem, your two middle numbers are 8 & 10. Since only one number can be the median, you add them together, and divide by 2:
8+10=18 18/2=9 < this is your middle, or MEDIAN
Next, your first and third quartiles-- they're the median of the lower half and data, and upper half.
For the lower quartile, find the mean of 2,4,5,6, and 8. again, cross out one from each side until you get to the middle. FIRST QUARTILE = 5
Do the same process for the upper half of data (10,13,17,19 & 20).
THIRD QUARTILE = 17
The MIN is the lowest number of data = 2
The MAX is the highest number of data = 20
Best of luck!
Answers:
a) 0.0625
b) 0.9375
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Work Shown:
The probability of landing on heads is 1/2 = 0.5 since both sides are equally likely to land on. Getting 4 heads in a row is (1/2)^4 = (0.5)^4 = 0.0625
The event of getting at least one tail is the complement of getting all four heads. This is because you either get all four heads or you get at least one tail. One or the other must happen. We subtract the result we got from 1 to get 1-0.0625 = 0.9375
You can think of it like this
P(getting all four heads) + P(getting at least one tail) = 1
The phrasing "at least one tail" means "one tail or more".
