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anonymous

  • one year ago

Amy has 35 tiles of the same shape and size in a box. The colors of and amounts of the tiles are shown below: 12 red tiles 8 yellow tiles 15 pink tiles Without looking in the box, Amy takes out a tile at random. She then replaces the tile and takes out another tile from the box. What is the probability that Amy takes out a yellow tile in both draws? 8 over 35 multiplied by 7 over 34 equal 56 over 1190 8 over 35 multiplied by 8 over 35 equal 64 over 1225 8 over 35 plus 7 over 34 equal 517 over 1190 8 over 35 plus 8 over 35 equal 560 over 1225

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. anonymous
    • one year ago
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    @Michele_Laino

  3. Michele_Laino
    • one year ago
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    we have 8 favorable outcomes and 12+8+15= 35 possible outcomes so probability= favorable outcomes/ possible outcomes=...

  4. Michele_Laino
    • one year ago
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    that is the probability associated to a single extraction

  5. Michele_Laino
    • one year ago
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    favorable outcomes = 8 possible outcomes = 35 \[\Large \frac{{{\text{favorable outcomes}}}}{{{\text{possible outcomes}}}} = ...\]

  6. anonymous
    • one year ago
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    so i have to divide

  7. Michele_Laino
    • one year ago
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    yes!

  8. anonymous
    • one year ago
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    @Michele_Laino

  9. Michele_Laino
    • one year ago
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    the probability is 8/35 it is simple!

  10. Michele_Laino
    • one year ago
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    for a single extraction

  11. anonymous
    • one year ago
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    so what about the other part

  12. Michele_Laino
    • one year ago
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    for second extraction, the probability is the same, namely it is 8/35, since Amy replaces the tile into the box. Now the requested probability is the product of the two probabilities, namely: probability = (8/35)*(8/35)=...?

  13. anonymous
    • one year ago
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    64/1225

  14. Michele_Laino
    • one year ago
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    that's right!

  15. anonymous
    • one year ago
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    Yay!

  16. Michele_Laino
    • one year ago
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    :)

  17. anonymous
    • one year ago
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    A coin was flipped 150 times. The results of the experiment are shown in the following table: Heads Tails 90 60 Which of the following best describes the experimental probability of getting heads? It is 10% higher than the theoretical probability. It is 10% lower than the theoretical probability. It is equal to the theoretical probability for this data. The experimental probability cannot be concluded from the data in the table.

  18. Michele_Laino
    • one year ago
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    the theoretical probability is 1/2 since we have two possible outcomes, namely a head and a tail and we have one favorable outcome, namely a head, so probability = 1/2= 0.5 The experimental probability to get a head is 90/150=...

  19. Michele_Laino
    • one year ago
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    since we have 90 favorable cases and 150 possible cases or outcomes

  20. anonymous
    • one year ago
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    0.6

  21. anonymous
    • one year ago
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    SO IT 10% lower than the theoretical probability

  22. Michele_Laino
    • one year ago
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    now the difference between the experimental probability and the theoretical probability is: 0.6-0.5=...

  23. anonymous
    • one year ago
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    so its 10% higher

  24. Michele_Laino
    • one year ago
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    that's right!

  25. anonymous
    • one year ago
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    so the answer is A

  26. Michele_Laino
    • one year ago
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    yes!

  27. anonymous
    • one year ago
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    yay!!!! last question sorry

  28. Michele_Laino
    • one year ago
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    ok!

  29. anonymous
    • one year ago
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    The number of marbles of different colors stored in a hat is listed below: 8 red marbles 10 green marbles 6 blue marbles Without looking in the hat, Tessa takes out a marble at random. She replaces the marble and then takes out another marble from the hat. What is the probability that Tessa takes out a blue marble in both draws? fraction 1 over 16 fraction 1 over 12 fraction 1 over 4 fraction 1 over 2

  30. Michele_Laino
    • one year ago
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    here we have 8+10+6= 24 possible outcomes and we have 6 favorable outcomes so the probability associated to a single extraction is: p= 6/24 please simplify that fraction

  31. anonymous
    • one year ago
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    0.25

  32. Michele_Laino
    • one year ago
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    we can divide both numerator and denominator by 6, wha do you get?

  33. anonymous
    • one year ago
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    divide what?

  34. anonymous
    • one year ago
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    oh 1/4

  35. Michele_Laino
    • one year ago
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    that's right!

  36. anonymous
    • one year ago
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    so is that the answer @Michele_Laino

  37. Michele_Laino
    • one year ago
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    now, the probability associated to the second extraction is the same, namely 1/4. So the requested probability is given by the product between the above probabilities, namely p= (1/4)*(1/4)=...?

  38. Michele_Laino
    • one year ago
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    what is: \[\Large P = \frac{1}{4} \times \frac{1}{4} = ...?\]

  39. anonymous
    • one year ago
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    Dang it

  40. anonymous
    • one year ago
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    i just submitted my test

  41. anonymous
    • one year ago
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    and i got it wrong

  42. anonymous
    • one year ago
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    because i forgot to do the last part

  43. anonymous
    • one year ago
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    ):

  44. anonymous
    • one year ago
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    but i got a 87% percent on my test

  45. Michele_Laino
    • one year ago
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    so have you passed it?

  46. anonymous
    • one year ago
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    yes i did

  47. Michele_Laino
    • one year ago
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    CONGRATULATIONS!!!!!

  48. anonymous
    • one year ago
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    im finished with math until next year yay!!!!!!!!

  49. Michele_Laino
    • one year ago
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    :)

  50. anonymous
    • one year ago
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    Im going to be in 8th grade

  51. Michele_Laino
    • one year ago
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    ok!! :)

  52. anonymous
    • one year ago
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    but Atleast i got a high B on my test !

  53. anonymous
    • one year ago
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    and i have a A in Math!

  54. Michele_Laino
    • one year ago
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    good job!!

  55. anonymous
    • one year ago
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    Thank you so much you really helped me undertand what i was doing. But i will be coming back just in case i need help lol!

  56. Michele_Laino
    • one year ago
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    ok! :)

  57. Jack1
    • one year ago
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    hi marc just if u could, please only post 1x question per post cheers

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