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jasenka [17]
3 years ago
6

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

Mathematics
1 answer:
stiks02 [169]3 years ago
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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Step 1

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6*6=36

Step 2

By using proportion find the percentage

\frac{150}{100} =\frac{36}{x}  \\ \\ 150*x=36*100\\\\ x= \frac{3600}{150} \\ \\ x=24

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the answer is

24%

3 0
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Step-by-step explanation:

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4 years ago
If r(x) = 3x – 1 and s(x) = 2x 1, which expression is equivalent to (startfraction r over s endfraction) (6)? startfraction 3 (6
viva [34]

The expression which is equivalent to the given ratio of the two provided binary fraction is startfraction 3 (6) minus 1 over 2 (6) 1 endfraction.

<h3>What is the equivalent expression?</h3>

Equivalent expressions are the expression whose result is equal to the original expression, but the way of representation is different.

The given binary functions are.

r(x) = 3x - 1

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The expression which has to be found out is,

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Thus, the expression which is equivalent to the given ratio of the two provided binary fraction is startfraction 3 (6) minus 1 over 2 (6) 1 endfraction.

Learn more about the equivalent expression here;

brainly.com/question/2972832

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For this problem you have to isolate the variable. To do this you must inverse the operations. Multiply 18 by 5, then divide by 12 to get your answer of 7.5
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