How many 12-digit decimal sequences are there that start and end with a sequence of at least two 3s?

1 answer:

6 0

Four positions of such a sequence are given to be 3s, and we have 10 choices for the remaining eight positions. So the number of such sequences is

$1^4\cdot10^8=10^8=100,000,000$

or 100 million.

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x² - 6x + 8 = 0
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When x = 2 , y = 2(2) - 4 = 0
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Points of intersection = A(2, 0) and B(4, 4)

Find the length of AB:

</span> $
\text {Length of AB}= \sqrt{(4-0)^2 + (4-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 4.47 \text{ units}$
<span>
Answer: 4.47 units

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