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8 June, 19:38

Carter took 7 coins out of his pocket. He did not have any pennies. He had more than $0.95 and less than $1.13 what could his coins have been? Prove your anser

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  1. 8 June, 21:37
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    Carter's coins could be;

    3 of 5¢, 2 of 10¢, 1 of 25¢, and 1 of 50¢

    Step-by-step explanation:

    The available coin denominations are;

    1¢ = 1 penny, 5¢, 10¢, 25¢, 50¢, and $1.00

    Since the coins are more than $0.95, we can have;

    5¢ * 3

    10¢ * 2

    25¢ * 1

    50¢

    Total = $1.00

    That is his coins could have been 3 * 5¢, 2 * 10¢, 1 * 25¢, and 1 * 50¢, to make a total of 7 coins with an amount value of $1.00.
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