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shusha [124]
3 years ago
12

Find solution to initial value problem: y'' +4y' +7y = 0; y(0) = 1, y'(0) = -2

Mathematics
1 answer:
rusak2 [61]3 years ago
4 0
Try this option, modify design according to local requirements.

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Drag the numbers below to put them in order from least to greatest: 4 16 1 7 -2 -20​
Mariulka [41]

Answer:

Are the 1 and 7 rogether? because if so then -20, -2, 4, 16, 17, if not then -20, -2, 1, 4, 7, 16

7 0
3 years ago
What is the solution to the system of equations?
frez [133]
(-19,55) x is equal to -19 and y is equal to 55
4 0
3 years ago
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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
The length of a rectangle is equal to triple the width. Write a system of equations can be used to tind the dimensions of the re
PtichkaEL [24]

Answer:

98 = 2w + 2L,

L = 3w

Step-by-step explanation:

Use the perimeter formula:

2) Perimeter = 2w + 2L (where w is the width and L is the length)

Since the length is three times the width we can write:

L = 3w

We can substitue L with 3w:

Perimeter is equal to 98 so,

98 = 2w + 6w = 8w

W = 98/8

Therefore w is 12.25, and L is 36.75

5 0
2 years ago
Will give brainliest!!
Dima020 [189]

Answer:

Sure this is a question from probability

Possible 2^n can be use

2^35 =3.432597387 × 10^10

Step-by-step explanation:

I'm not sure

6 0
3 years ago
Read 2 more answers
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