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Rzqust [24]
2 years ago
15

The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i

n the $n^{\rm{th}}$ block. What is the sum of the first $1234$ terms of this sequence
Mathematics
1 answer:
kicyunya [14]2 years ago
8 0

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

numbers, and their sum is

\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

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A shipment of 50,000 transistors arrives at a manufacturing plant. The quality control engineer at the plant obtains a random sa
Aleks04 [339]

Step-by-step explanation:

remember, the number of possible combinations to pick m out of n elements is C(n, m) = n!/(m! × (n-m)!)

50,000 transistors.

4% are defective, that means 4/100 = 1/25 of the whole.

so, the probability for one picked transistor to be defective is 1/25.

and the probability for it to work properly is then 1-1/25 = 24/25.

now, 500 picks are done.

to accept the shipment, 9 or less of these 500 picks must be defective.

the probability is then the sum of the probabilities to get

0 defective = (24/25)⁵⁰⁰

1 defective = (24/25)⁴⁹⁹×1/25 × C(500, 1)

= 24⁴⁹⁹/25⁵⁰⁰ × 500

2 defective = (24/25)⁴⁹⁸×1/25² × C(500, 2)

= 24⁴⁹⁸/25⁵⁰⁰ × 250×499

3 defective = 24⁴⁹⁷/25⁵⁰⁰ × C(500, 3) =

= 24⁴⁹⁷/25⁵⁰⁰ × 250×499×166

...

9 defective = 24⁴⁹¹/25⁵⁰⁰ × C(500, 9) =

= 24⁴⁹¹/25⁵⁰⁰ × 500×499×498×497×496×495×494×493×492×491 /

9×8×7×6×5×4×3×2 =

= 24⁴⁹¹/25⁵⁰⁰ × 50×499×166×71×31×55×494×493×41×491

best to use Excel or another form of spreadsheet to calculate all this and add it all up :

the probability that the engineer will accept the shipment is

0.004376634...

which makes sense, when you think about it, because 10 defect units in the 500 is only 2%. and since the whole shipment contains 4% defect units, it is highly unlikely that the random sample of 500 will pick so overwhelmingly the good pieces.

is the acceptance policy good ?

that completely depends on the circumstances.

what was the requirement about max. faulty rate in the first place ? if it was 2%, then the engineer's approach is basically sound.

it then further depends what are the costs resulting from a faulty unit ? that depends again on when the defect is usually found (still in manufacturing, or already out there at the customer site, or somewhere in between) and how critical the product containing such transistors is. e.g. recalls for products are extremely costly, while simply sorting the bad transistors out during the manufacturing process can be rather cheap. if there is a reliable and quick process to do so.

so, depending on repair, outage and even penalty costs it might be even advisable to have a harder limit during the sample test.

in other words - it depends on experience and the found distribution/probability curve, standard deviation, costs involved and other factors to define the best criteria for the sample test.

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X is from a to b translates to a≤x≤b<br><br> A. True<br> B. False
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Answer:

True

Step-by-step explanation:

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