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butalik [34]
2 years ago
4

Which recursive formula can be used to generate the sequence shown, where f(1) = 9.6 and n > 1? 9.6, –4.8, 2.4, –1.2, 0.6, ..

.
Mathematics
2 answers:
USPshnik [31]2 years ago
8 0
F(10 +5 >1 - 2)

With the domain restrictions you can solve and get your answers
muminat2 years ago
7 0

Answer:

Recursive formula will be T_{n}=9.6(0.5)^{n-1}

Step-by-step explanation:

The given sequence is 9.6, -4.8, 2.4, -1.2, 0.6......

In this sequence we find that there is a common ratio which makes this sequence a geometric sequence.

For a geometric sequence explicit formula is given by

T_{n}=a(r)^{n-1} if n > 1

So by putting values of a = 9.6

and common ratio r = \frac{-(4.8)}{9.6} = -\frac{1}{2}

Therefore, the recursive formula will be

T_{n}=9.6(0.5)^{n-1}

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Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

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For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

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We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable N_1 like this:

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And if we want to find the joint probability we just need to do this:

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And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

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2 years ago
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