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Alexxx [7]
2 years ago
14

A stick of length l is broken at a uniformly chosen random location. We denote the length of the smaller piece by X. (a) Find th

e cumulative distribution function of X. (b) Find the probability density function of X.
Mathematics
1 answer:
kykrilka [37]2 years ago
6 0

Answer:

a) \phi (x) = a/(I/2)

b) f(x) = 2/I

Step-by-step explanation:

a) Lets denote \phi the cumulative distribution function of X. Note that for any value a between 0 and I/2, we have that \phi(a) is the probability for the stick to be broken before the length a is reached following the stick from one starting point plus the probability for the stick to be broken after the length I-a from the same starting point. This means that \phi(a) = (a+a)/I = 2a/I = a/ (I/2)

b) Note that, as a consecuence of what we calculate in the previous item, X has a uniform distribution with parameter I/2, therefore, the probability density function f is

f(x) = 1/(I/2) = 2/I  

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Which statement about the simplified binomial expansion of (a + b)", where n is a positive integer, is true?
Alja [10]

Answer:

(a+b)^n  ={n \choose 0}a^{(n)}b^{(0)} + {n \choose 1}a^{(n-1)}b^{(1)} + {n \choose 2}a^{(n-2)}b^{(2)} + .....  +{n \choose n}a^{(0)}b^{(n)}

Step-by-step explanation:

The Given question is INCOMPLETE as the statements are not provided.

Now, let us try and solve the given expression here:

The given expression is: (a +b)^n, n > 0

Now, the BINOMIAL EXPANSION is the expansion which  describes the algebraic expansion of powers of a binomial.

Here, (a+b)^n  = \sum_{k=0}^{n}{n \choose k}a^{(n-k)}b^{(k)}

or, on simplification, the terms of the expansion are:

(a+b)^n  ={n \choose 0}a^{(n)}b^{(0)} + {n \choose 1}a^{(n-1)}b^{(1)} + {n \choose 2}a^{(n-2)}b^{(2)} + .....  +{n \choose n}a^{(0)}b^{(n)}

The above statement holds for each  n > 0

Hence, the complete expansion for the given expression is given as above.

4 0
2 years ago
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9x-(-2y)-8x-(-12y) simplify​
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Answer:

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Step-by-step explanation:

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Lavania is studying the growth of a population of fruit flies in her laboratory. After 6 days she had nine more than five times
MatroZZZ [7]

Answer:

Lavania observed 39 fruit flies after 6 days of observation

Step-by-step explanation:

Let x be the number of fruit flies on the first day of Lavania's study.

After 6 days she had nine more than five times as many fruit flies as when she began the study.

Five times as many fruit flies as when she began the study = 5x

Nine more than five times as many fruit flies as when she began the study=5x+9

The expression to find the population of fruit flies Lavania observed after 6 days is 5x+9

If she observes 20 fruit flies on the first day of the study, then x=6, then

5x+9=5\cdot 6+9=30+9=39

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Answer:

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3)x = 7

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