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Hypergeometric Distribution
Consider the previous example of the batch of light bulbs. Suppose the Bernoulli experiment is repeated without replacement. That is, once a bulb is tested, it will be set aside.
If X = number of successes, then
It can be shown that for the Hypergeometric Distribution
It should be intuitively clear that if nM/N is small enough, the hypergeometric distribution may be replaced by the binomial distribution. In our development of the subject, we will do so if n/N £ 0.05.
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Fermat's Theorem If f(x) has a relative extrema at x = c and f′(c) exists then x = c is a critical point of f(x). Actually, this will be a critical point that f′(c) =0.
Find the circumference of a circle whose area is 16 times the area of the circle with diameter 7cm (Ans: 88cm) Ans: Π R 2 = 16 Π r 2 R 2 = 16 r 2
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Illustrates that the following numbers aren't solutions to the given equation or inequality. y = -2 in 3( y + 1) = 4 y - 5 Solution In this case in essence we do the sam
how do you slove 4u-5=2u-13
It is the last case that we require to take a look at. During this section we are going to look at solutions to the system, x?' = A x? Here the eigenvalues are repeated eigen
Explain Factor by Grouping ? Factoring by grouping is often a good way to factor polynomials of 4 terms or more. (Sometimes it isn't. It doesn't always work. But it's worth try
Determine the second derivative for following functions. Q (t ) = sec (5t ) Solution : Following is the first derivative. Q′ (t
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