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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

angle BCE = angle BCA + angle

**ACE**Let's find angle BCA and angle**ACE**: angle**ACE**...1 Answers · Science & Mathematics · 18/05/2019

...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

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

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

... 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

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

...? 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

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

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) = 396/26522 Answers · Science & Mathematics · 20/01/2018

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