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Verdich [7]
2 years ago
9

A number is chosen uniformly at random from among the positive integers less than $10^8$. Given that the sum of the digits of th

e number is 9, what is the probability that the number is prime
Mathematics
1 answer:
Reil [10]2 years ago
3 0

Answer:

 

If the sum of the digits of a number is divisible by 9, then that number is divisible by 9. An..... sorry if this does not help :(

Step-by-step explanation:

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
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f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
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\\ \quad \\
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\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
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\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
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\bullet \textit{ vertical shift by }{{  D}}\\
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so hmm if you notice, your f(x) there, looks just like g(x), but is "shifted horizontally" by about 9 units to the right

that simply means, based on that template above, that C/B = -9, and we can simply make B = 1  and C = -9, and we get C/B = -9/1 which is -9

thus   \bf \begin{array}{llcll}
f(x)=g(&1x&-9)\\
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Answer:

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Step-by-step explanation:

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Let's check each case to determine the solution to the problem.

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