Two cards are chosen at random without replacement what is the probability of getting two aces

Mathematics

1 answer:

VARVARA [1.3K]3 years ago

4 0

4/52 chance of getting an ace.

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Which situation could be represented by the equation - 15 + 15 = 0?

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If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$