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anonymous

  • one year ago

WILL MEDAL AND FAN NO LIE Cards numbered 1, 1, 2, 3, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, and 9 are placed in a bag. If a card is drawn from the bag and then replaced 135 times, about how many times would a 6 be drawn? A. 2 B. 9 C. 15 D. 18

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  1. Vocaloid
    • one year ago
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    (number of 6's/total number of cards)*135

  2. anonymous
    • one year ago
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    im confused

  3. Vocaloid
    • one year ago
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    how many cards have 6 on them?

  4. welshfella
    • one year ago
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    there are 2 cards marked 6 out of a total of 15 cards so probability of picking a 6 in one draw is = ?

  5. anonymous
    • one year ago
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    2 @Vocaloid

  6. Vocaloid
    • one year ago
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    right, and how many total cards are there?

  7. anonymous
    • one year ago
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    @welshfella

  8. welshfella
    • one year ago
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    I mentioned that above

  9. anonymous
    • one year ago
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    IS IT C

  10. anonymous
    • one year ago
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    @welshfella

  11. Vocaloid
    • one year ago
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    the answer isn't c, would you mind telling me how many total cards there are?

  12. anonymous
    • one year ago
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    wait i think its D srry

  13. jim_thompson5910
    • one year ago
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    As Vocaloid wrote \[\Large \text{Expected Number of 6's} = \frac{\text{Number of 6's}}{\text{Total number of cards}}*135\] \[\Large \text{Expected Number of 6's} = \frac{2}{15}*135\] \[\Large \text{Expected Number of 6's} = ???\]

  14. anonymous
    • one year ago
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    sooo just do 2/15 x 135 :/

  15. jim_thompson5910
    • one year ago
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    yes

  16. anonymous
    • one year ago
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    ok

  17. anonymous
    • one year ago
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    18 soooo D

  18. anonymous
    • one year ago
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    @jim_thompson5910

  19. jim_thompson5910
    • one year ago
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    yep 18

  20. anonymous
    • one year ago
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    ok can u help me with more @jim_thompson5910

  21. anonymous
    • one year ago
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    @automaticloveletter

  22. anonymous
    • one year ago
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    @taramgrant0543664

  23. anonymous
    • one year ago
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    HELP PLZZZZ

  24. taramgrant0543664
    • one year ago
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    Is there a new question?

  25. anonymous
    • one year ago
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    yea i need help with math

  26. taramgrant0543664
    • one year ago
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    Well you can ask them and just tag me in them I'm happy to help

  27. anonymous
    • one year ago
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    ok

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