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4 years ago

11

There are 2 green marbles, 7 blue marbles, 3 red marbles, and 4 teal marbles in a jar. If two marbles are selected and not repla

ced, what is the probability of selecting a blue marble and a teal marble?

1 answer:

Anna007 [38]4 years ago

3 0

Step-by-step explanation:

Hello there!

Multiply the probabilitites of the two outcomes:

7/16 x 1/4=7/64

There you have it.

Have a great day!

:)

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3 years ago

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Following are the given series for all x:

Step-by-step explanation:

Given equation:

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Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

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$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

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