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Akimi4 [234]
3 years ago
6

Internet providers: For the data in Exercise 48, estimate the number of Internet plans whose cost is

Mathematics
1 answer:
grandymaker [24]3 years ago
3 0
The answer is c , between 52.43 and 64.20
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Write the sum using summation notation, assuming the suggested pattern continues.100 + 121 + 144 + 169 + ... + n2 + ...
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Sigma_n=10 ^infinity n^2


It would be infinite.



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HELP ME PLEASE I"M FAILING
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A blueprint of a sailboat has a sail in the shape of a
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Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of
steposvetlana [31]

Cards are drawn, one at a time, from a standard deck; each card is replaced before the next one is drawn. Let X be the number of draws necessary to get an ace. Find E(X) is given in the following way

Step-by-step explanation:

  • From a standard deck of cards, one card is drawn. What is the probability that the card is black and a jack? P(Black and Jack)  P(Black) = 26/52 or ½ , P(Jack) is 4/52 or 1/13 so P(Black and Jack) = ½ * 1/13 = 1/26
  • A standard deck of cards is shuffled and one card is drawn. Find the probability that the card is a queen or an ace.

P(Q or A) = P(Q) = 4/52 or 1/13 + P(A) = 4/52 or 1/13 = 1/13 + 1/13 = 2/13

  • WITHOUT REPLACEMENT: If you draw two cards from the deck without replacement, what is the  probability that they will both be aces?

P(AA) = (4/52)(3/51) = 1/221.

  • WITHOUT REPLACEMENT: What is the probability that the second card will be an ace if the first card is a  king?

P(A|K) = 4/51 since there are four aces in the deck but only 51 cards left after the king has been  removed.

  • WITH REPLACEMENT: Find the probability of drawing three queens in a row, with replacement. We pick  a card, write down what it is, then put it back in the deck and draw again. To find the P(QQQ), we find the

probability of drawing the first queen which is 4/52.

  • The probability of drawing the second queen is also  4/52 and the third is 4/52.
  • We multiply these three individual probabilities together to get P(QQQ) =
  • P(Q)P(Q)P(Q) = (4/52)(4/52)(4/52) = .00004 which is very small but not impossible.
  • Probability of getting a royal flush = P(10 and Jack and Queen and King and Ace of the same suit)
5 0
3 years ago
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