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Yuri [45]
3 years ago
14

An integer from 100 through 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have

0 as at least 1 digit?
Mathematics
1 answer:
Nat2105 [25]3 years ago
6 0
So, we know that for ever 100 digits, there are 10 digits that have a 0 in the one's place (100, 110, 120, 130, 140, 150, 160, 170, 180, and 190), and 9 different numbers with a 0 in the ten's place (101, 102, 103, 104, 105, 106, 107, 108, and 109). You do not count any number twice. That's 19/100. Now, that's an answer for ever 100, until you get to the 900's, since there's only 99 in that place.

We know there are 899 options, so that's our denominator. Then, 19+19+19+19+19+19+19+19+19=171.

The answer is 171/899, I believe.
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Read 2 more answers
How do I solve this question?
zubka84 [21]

Answer:

f(x) = \frac{108}{7} (\frac{7}{6})^x or f(x) = 15.429(1.167)^x

Step-by-step explanation:

No need for a calculator.

Assuming that the equation is in the form y = ab^x, we can plug in points to get our equation. For ease, let's use (1,18) and (2,21). When plugging these points in, we get 18 = ab^1, 21 = ab^2. Now let's divide the equations to get rid of a: \frac{21 = ab^2}{18 = ab^1}  = \frac{21}{18}  =  \frac{7}{6}  = b. Now that we have b, we can plug in the value we just calculated to solve for a: 18 = a(\frac{7}{6} )^1, and solving for a, we get a = \frac{108}{7}.

So the f(x) = (\frac{108}{7})(\frac{7}{6}  )^x

This equation in decimal form (rounded to the nearest thousandth) is f(x) = 15.429(1.167)^x.

hope this helped! :)

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Which of the following is equivalent to the expression (3ab)(-5ab)?
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