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statuscvo [17]
3 years ago
10

Pls help....which of them is it and no link or I will report​

Mathematics
1 answer:
Serga [27]3 years ago
3 0
Answer W = 39/4
9 would be the answer
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2x+5=10<br> I need to find out x pls help​
Finger [1]

Answer:

5

Step-by-step explanation:

2x + 5 = 10

you minus the 5

2x = 5

you divide both sides by 2

x = 5

7 0
3 years ago
Read 2 more answers
Anyone can help? with this problem
Iteru [2.4K]

Answer:

E

Step-by-step explanation:

24 is 4 in the ratio so 24÷4 gives 6 and 5x6 is 30 so the hypotenuse is 30 inches

4 0
3 years ago
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
For cones with radius 6 units, the equation V = 127th relates the height h of the cone, in units, and the volume V of the cone,
denis-greek [22]

Answer:

hola

Step-by-step explanation:

cómo estás yo bien compita

6 1
2 years ago
Read 2 more answers
PLease give me answer ASAP
fomenos

Answer:

             

Step-by-step explanation:

7 0
3 years ago
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