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Paraphin [41]
4 years ago
8

From a pack of 9 cards numbered 1-9, three cards are drawn at random and laid on a table from left to right.

Mathematics
1 answer:
Andreas93 [3]4 years ago
4 0
Answer: 1/6
Explanation: Since we need to draw three cards from a pack of 9 and the digits require an order, then our sample space becomes: 9P3.
Now, we don't care about picking an order when picking three cards to analyse, as there is only 1 way in arranging the three cards in descending order.

Thus, we get:

\frac{^{9}C_3}{^{9}P_3} = \frac{1}{6}
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Please help. I don't understand what to input into the c...
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Factorize the denominator:

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Then we find that ...

• When <em>c</em> = 15,

\displaystyle \lim_{x\to15}f(x) = \lim_{x\to15}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to15}\frac1{x-16} = \frac1{15-16} = \frac1{-1} = \boxed{-1}

because the factors of <em>x</em> - 15 in the numerator and denominator cancel with each other. More precisely, we're talking about what happens to <em>f(x)</em> as <em>x</em> gets closer to 15, namely when <em>x</em> ≠ 15. Then we use the fact that <em>y</em>/<em>y</em> = 1 if <em>y</em> ≠ 0.

• When <em>c</em> = 16,

\displaystyle \lim_{x\to16}f(x) = \lim_{x\to16}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to16}\frac1{x-16} = \dfrac10

which is undefined; so this limit does not exist.

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\displaystyle \lim_{x\to17}f(x) = \lim_{x\to17}\frac{x-15}{(x-16)(x-15)} = \lim_{x\to17}\frac1{x-16} = \frac1{17-16}=\frac11 =\boxed{1}

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