5,5 liter : ½ liter = tolong di jawab

In the first 10 -> 10 - 9 = 1 (contain a 3). In the first 100 -> 100 - 9 * 9 = 19 (contain a 3). In the first 10^n ->** 10^n - 9^n** (contain a 3).
[
So for 10^4 -> 10^4 - 9^4 = 10.000 - 6.561 = 3.439 ]
The answer is 3.439 numbers contain a 3 in the first 10.000

Answered by Anonymous | 2024-06-10

we know that
It is much easier to compute the number of numbers that do not contain a 3 .
In the first 10 numbers 9 do not contain a 3 .
In the first 100 numbers 81 do not contain a 3 . (This is because these numbers are formed by selecting a digit out of
[ 0 , 1 , 2 , 4 , 5 , 6 , 7 , 8 , 9 ] for the first position and another digit out of the same set for the last position and this may be done in
9 ∗ 9 = 81 w a ys .
In the first 1000 numbers, reasoning as above, there are
9 3 numbers that do not contain a 3 .
Now the pattern should be becoming obvious.
In the first 1 0 n numbers there are 9 n numbers that do not contain a 3 .
So
In the first 1 0 4 numbers there are 9 4 numbers that do not contain a 3 .
9 4 = 6 , 561
10 , 000 − 6 , 561 = 3 , 439
Thus
the ratio of those numbers that do contain a 3 over the total is equal to
R a t i o = 10 , 000 3 , 439 ​ R a t i o = 0.3439
therefore
the answer is
0.3439

Answered by calculista | 2024-06-11

The proportion of the first 10,000 natural numbers that contain the digit 3 is approximately 0.4139. We calculated this by first determining how many numbers do not contain a 3 and subtracting that from 10,000. Ultimately, this calculation shows that about 41.39% of these numbers include the digit 3.
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Answered by calculista | 2024-09-05

[tex]5.5 \div \frac{1}{2} \\ \\ = \frac{55}{10} \times \frac{2}{1} \\ \\ = \frac{110}{10} = 11 \: \: liter[/tex]

Answered by aquila0152 | 2025-07-04