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25 January, 04:16

Teo has a combination of quarters, loonies, and toonies in his pocket.

He knows he has 9 coins in total and that he has at least one of each

type of coin.

a) What is the probability that Teo has more than $8?

b) How many different combinations of coins could Teo use to pay for an item

that costs between $10 and $12, if he uses fewer toonies than loonies?

+3
Answers (2)
  1. 25 January, 04:43
    0
    343

    Step-by-step explanation:

    There are quarters (25¢), loonies ($1), and toonies ($2)

    And total number of coins is 9

    If loonies ($1), and toonies ($2) are one, then quarters (25¢) is 7

    If quarters (25¢), and toonies ($2) are one, then loonies ($1) is 7

    If loonies ($1), and quarters (25¢) are one, then toonies ($2) is 7

    So possible number number of combination for an item that cost between $10 and $12 is

    = 1^3 * 1^3 * 7^3

    = 343
  2. 25 January, 04:46
    0
    ooga booga ooga booga ooga booga ooga booga
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