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Fantom [35]
3 years ago
9

Suppose that you have 12 identical balls and 3 empty boxes. You randomly put every ball into some box. What is the probability t

hat none of the boxes have more than 6 balls?
Mathematics
1 answer:
lawyer [7]3 years ago
7 0

Answer:

the probability that none of the boxes have more than 6 balls is 0.3077

Step-by-step explanation:

Given that;

12 balls are put into 3 boxes randomly, without ay condition

so we will be using the multinomial formula;

⇒ [ 12 + 3 - 1              [ 14

        3 - 1 ]         =       2 ]    =    91

now, assuming that one of the box has more than 6 balls that is at least 7 balls

x + y + z = 12

 x + y + z = 7

therefore

x + 7 + y + z = 12

x + y + z = 5

therefore the number of the solution here computed as;

⇒ [ 5 + 3 - 1              [ 7

        3 - 1 ]         =       2 ]    =   21

Hence, the probability that none of the boxes have more than six (6) balls will be;

= (91 - (3 × 21)) / 91

= (91 - 63) / 91

= 28 / 91

= 0.3077

Therefore the probability that none of the boxes have more than 6 balls is 0.3077

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For the function g(x)=3-8(1/4)^2-x
Reika [66]

Using function concepts, it is found that:

  • a) The y-intercept is y = 2.5.
  • b) The horizontal asymptote is x = 3.
  • c) The function is decreasing.
  • d) The domain is (-\infty,\infty) and the range is (-\infty,3).
  • e) The graph is given at the end of the answer.

------------------------------------

The given function is:

g(x) = 3 - 8\left(\frac{1}{4}\right)^{2-x}

------------------------------------

Question a:

The y-intercept is g(0), thus:

g(0) = 3 - 8\left(\frac{1}{4}\right)^{2-0} = 3 - 8\left(\frac{1}{4}\right)^{2} = 3 - \frac{8}{16} = 3 - 0.5 = 2.5

The y-intercept is y = 2.5.

------------------------------------

Question b:

The horizontal asymptote is the limit of the function when x goes to infinity, if it exists.

\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2+\infty} = 3 - 8\left(\frac{1}{4}\right)^{\infty} = 3 - 8\frac{1^{\infty}}{4^{\infty}} = 3 -0 = 3

--------------------------------------------------

\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 3 - 8\left(\frac{1}{4}\right)^{2-x} = 3 - 8\left(\frac{1}{4}\right)^{2-\infty} = 3 - 8\left(\frac{1}{4}\right)^{-\infty} = 3 - 8\times 4^{\infty} = 3 - \infty = -\infty

Thus, the horizontal asymptote is x = 3.

--------------------------------------------------

Question c:

The limit of x going to infinity of the function is negative infinity, which means that the function is decreasing.

--------------------------------------------------

Question d:

  • Exponential function has no restrictions in the domain, so it is all real values, that is (-\infty,\infty).
  • From the limits in item c, the range is: (-\infty,3)

--------------------------------------------------

The sketching of the graph is given appended at the end of this answer.

A similar problem is given at brainly.com/question/16533631

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Answer:

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