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Math

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A speaker has a collection of 62 jokes. He uses 3 jokes in each speech.

Please help i got a is 37820 but really need help with part b

(a) In how many ways can he select 3 jokes for a speech?

How many speeches can he give without using a joke twice? (Round your answer down to the nearest whole number.)

  • Math -

    a) is correct

    b) If I interpret it correctly

    first speech: he has 62 jokes available
    second speech: he has 59 jokes available
    third speech: 56 jokes
    ....
    nth speech: ≥ 3

    so we have to find the number of terms in the arithmetic sequence
    62, 59, 56 , .... , 3 or 2 or 1
    a = 62, d=-3 , n = ? when term(n) = 3
    a
    term(n) = a+(n-1)d
    3 = 62 + (n-1)(-3
    -59 = -3n + 3
    3n = 62
    n = 62/3 = 20.666..

    check:
    term(21) = 62 -3(20) = 2
    term(20) = 62 - 3(19) = 5

    So he should be able to give 20 speeches, but on the 21st speech he would only have 2 jokes to tell

  • Math -

    I understand what you did but which one would i round it to the 20 speeches or 21 speeches?

  • Math -

    Well, your instructions were to round down to the nearest whole number.
    As I said, for his 20th speech he would have 5 jokes left that he hasn't used yet, and he chooses 3 of those.
    So on his 21st speech he only has 2 unused jokes, and would have to repeat one of the other jokes.

    So without repeating any jokes, my answer would be 20 speeches.

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