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3 years ago

Simplify 1/9-1/x 1/81-1/x^2

Mathematics

1 answer:

tekilochka [14]3 years ago

5 0

$\cfrac{ \frac{1}{9}- \frac{1}{x} }{ \frac{1}{81} - \frac{1}{x^2} }= \cfrac{ \frac{x-9}{9x} }{ \frac{x^2-81}{81x^2} }= \cfrac{x-9}{9x}* \cfrac{81x^2}{(x-9)(x+9)}= \cfrac{9x}{x+9}$

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The correct answer is 0.94147

Step-by-step explanation:

Let A denote the event that the podiatrist finds the first person with an ingrown toenail.

And (1 - A) denote the event that the podiatrist does not find the ingrown toenail.

While examining seven people, the podiatrist can find the very first person to have an ingrown toenail. Similarly he can find the second patient to have the ingrown toenail. Going in this way the probability of the first person to have an ingrown toenail is given by:

= A + (1 - A) × A + (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × (1 - A) × (1 - A) × (1 - A) × A.

= $\frac{1}{3} + \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} .$

= $\frac{1}{3} + \frac{2}{3} \frac{1}{3} + (\frac{2}{3}) ^{2} \frac{1}{3} + (\frac{2}{3})^{3} \frac{1}{3} + (\frac{2}{3})^{4} \frac{1}{3} + (\frac{2}{3})^{5} \frac{1}{3} + (\frac{2}{3})^{6} \frac{1}{3}.$

= 0.94147

We can also solve the above expression by using the geometric progression formula as well where common ratio is given by $\dfrac{2}{3}$ .

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