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Sindrei [870]
3 years ago
13

Help Please!!

Mathematics
1 answer:
Finger [1]3 years ago
6 0
Answer is A. r must be such that
-1 ≤ r ≤ 1
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What is the GCF of 28x3 and 28x5?
Sergio039 [100]

Answer:

28x^3

Step-by-step explanation:

8 0
3 years ago
CAN ANYONE HELP ME ANSWER THIS ​
azamat

Answer:

9/100

Step-by-step explanation:

There are 10 marbles in the bag

There are 3 pink marbles.

P (1st marble pink) = pink marbles/ total marbles

                              = 3/10

We put the marble back.

There are 10 marbles in the bag

There are 3 pink marbles.

P (2nd marble pink) = pink marbles/ total marbles

                                      = 3/10

Multiply the probabilities together

P (pink, then pink) = 3/10 * 3/10 = 9/100

7 0
3 years ago
Solve the inequality. Draw a number line representing the solution-<br> -5x + 12 &gt;7
andrew-mc [135]
Ok the answer is X < 1
3 0
3 years ago
A parabola has two x intercepts at (-2 , 0) and (3 , 0) and passes through the point (5 , 10). Find the equation of this parabol
Kamila [148]
Y = a(x + 2)(x - 3) : equation of the parabola in factored form 10 = a(5 + 2)(5 - 2) : (5 , 10) is a point on the graph of the parabola and therefore satisfies the equation of the parabola. a = 5/7: solve the above equation for a.
4 0
3 years ago
A card is chosen from a standard deck of cards. What is the probability that the card is a club, given that the card is black?
leonid [27]
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. 1 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) What's the probability of being dealt a royal flush in a five card hand from a standard deck of cards? (Note: A royal flush is a 10, Jack, Queen, King, and Ace of the same suit. A standard deck has 4 suits, each with 13 distinct cards, including these five above.) (NB: The order in which the cards are dealt is unimportant, and you keep each card as it is dealt -- it's not returned to the deck.) The probability of drawing any card which could fit into some royal flush is 5/13. Once that card is taken from the pack, there are 4 possible cards which are useful for making a royal flush with that first card, and there are 51 cards left in the pack. therefore the probability of drawing a useful second card (given that the first one was useful) is 4/51. By similar logic you can calculate the probabilities of drawing useful cards for the other three. The probability of the royal flush is therefore the product of these numbers, or 5/13 * 4/51 * 3/50 * 2/49 * 1/48 = .00000154
5 0
3 years ago
Read 2 more answers
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