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  1. One way to keep order from counting is to only count sequences that are sorted. That can help in developing an algorithm to find counts for each possible sum. Define N(n, t, s) to be the number of sorted...

    1 Answers · Science & Mathematics · 09/10/2019

  2. angle BCE = angle BCA + angle ACE Let's find angle BCA and angle ACE: angle ACE...

    1 Answers · Science & Mathematics · 18/05/2019

  3. ...and their IQ score. But it's still possible for someone to ace the mathematical questions on test and get a lower IQ score...

    7 Answers · Science & Mathematics · 07/05/2019

  4. You have a King , so there are C(51,4) ways for other 4 cards out of 51 cards . But these 4 cards must be (A, Q, J, 10) or (Q, J, 10, 9) to complete a 5-card straight hand . If these 4 cards...

    3 Answers · Science & Mathematics · 17/07/2018

  5. Set up a proportion: BD/CE = AD/AE 5.0 / 7.4 = x / (x+2.7) Cross multiply: 7.4x = 5.0(x + 2.7) 7.4x = 5.0x + 13.5 2.4x = 13.5 x = 13.5/2.4 x = 135/24 x = 45/8 x = 5 5/8 x = 5.625 x ≈ 5.6 cm

    1 Answers · Science & Mathematics · 23/09/2018

  6. ... are 4 Kings in the deck * There are 4 Aces in the deck Under simple probability with order (King first, then...

    1 Answers · Science & Mathematics · 08/05/2018

  7. Have you tried going back to a question on an exam that flustered you to try and figure out what caused that? Can you perhaps see that it wasn't that different but just that you were scared about it? One thing may be that at the high school level, we...

    2 Answers · Science & Mathematics · 13/03/2018

  8. ...? Pair of cards: 14 It's entirely possible to draw an Ace, 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , Jack , Queen, and King...

    3 Answers · Science & Mathematics · 13/02/2018

  9. This is conditional probability. It says, "if you draw a heart, what is the probability it is the queen of hearts?" There are 13 heart cards. You don't have to consider the other 39 cases because you are...

    6 Answers · Science & Mathematics · 10/02/2018

  10. P(at least 1 Ace) = 1 - P(both no aces) P(1st card no ace) = number...51 P(both no aces) = 48/52 * 47/51 = 2256/2652 P(at least 1 Ace) = 1 - P(both no aces) = 1 - (2256/2652) = 396/2652

    2 Answers · Science & Mathematics · 20/01/2018

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