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tatyana61 [14]
4 years ago
13

You flip a fair coin 10,000 times. Approximate the probability thatthe difference between the number of heads and the number of

tails is at most 100.
Mathematics
1 answer:
Harlamova29_29 [7]4 years ago
7 0

Answer:

E[X] = (np) (p+(1-p))^(n-1)

With n =10,000 and p=0.50 we get

E[X]= (10,000*0.50) (0.50+(1–0.50))^(10,000–1)

E[X] = 5,000(1)

So assuming the coin is fair (p=50%), then we can expect to get heads 5,000 times when the coin is tossed 10,000 times.

Step-by-step explanation:

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<u>Pre-Algebra</u>

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<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Identify</u>

<em>a</em> = x + 3

<em>b</em> = x

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<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute [PT]:                                                                                                (x + 3)² + x² = (√117)²
  2. Expand [FOIL]:                                                                                                 x² + 6x + 9 + x² = (√117)²
  3. Combine like terms:                                                                                       2x² + 6x + 9 = (√117)²
  4. Exponents:                                                                                                      2x² + 6x + 9 = 117
  5. [SPE] Subtract 117 on both sides:                                                                    2x² + 6x - 108 = 0
  6. Factor out GCF:                                                                                               2(x² + 3x - 54) = 0
  7. [DPE] Divide 2 on both sides:                                                                         x² + 3x - 54 = 0
  8. Factor Quadratic:                                                                                            (x - 6)(x + 9) = 0
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Since we are dealing with positive values, we can disregard the negative root.

∴ x = 6

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