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Shtirlitz [24]
3 years ago
13

How many ways are there for Alice, Bob, Charlie, David and Eve to line up from left to right, such that Charlie is somewhere to

the left of Eve (but not necessarily immediately to the left)
Mathematics
1 answer:
steposvetlana [31]3 years ago
6 0

Answer:

120 ways

Step-by-step explanation:

Alice, Bob, David, Charlie,   Eve

David, Alice, Bob,  Charlie,   Eve

Alice,David, Bob,  Charlie,   Eve

Charlie,   Eve ,Alice,David, Bob,  

Charlie,   Eve ,Alice, Bob, David,

Charlie,   Eve ,David, Alice, Bob,

Alice,Charlie,   Eve , Bob, David,

Alice,Charlie,   Eve ,  David, Bob,

Bob,Alice,Charlie,   Eve ,  David,   and so .

This is a permutation question as the order of placing Charlie to the left of Eve is important.

So the total number of people n= 5 and the possible order is 4 keeping Charlie left of Eve. Eve cannot have the last position to keep Charlie on the left.

Using the formula of nPr = n!/ (n-r)! we get

5! / (5-4)! = 120 ways in which Charlie can be placed to the left of Eve.

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Chebyshev’s Theorem establishes that at least 1 - 1/k²  of the population lie among k standard deviations from the mean.

This means that for k = 2,  1 - 1/4 = 0.75. In other words, 75% of the total population would be the percentage of healthy adults with body temperatures that are within 2standard deviations of the mean.

The maximum value of that range would be simply  μ + 2s, where μ is the mean and s the standard deviation. In the same way, the minimum value would be μ - 2s:

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Step-by-step explanation:

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