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krek1111 [17]
3 years ago
9

Help me people............

Mathematics
1 answer:
lubasha [3.4K]3 years ago
8 0
 id go with line T if it was me
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Please help me I don’t understand.
attashe74 [19]

Answer:

1 : 64

Step-by-step explanation:

Model:

1/2 in * 1/6 in = 1/12 in^2

Real:

1/3 ft * 1/9 ft = 1/3 * 12 in * 1/9 * 12 in = 16/3 in^2

model : real

1/12 / 16/3 = 1/12 * 3/16 = 3/192 = 1/64

model : real = 1 : 64

7 0
3 years ago
Dennis needs to fix a leaky roof on his house but does not own a ladder. He thinks that a 25-foot ladder will be
Mars2501 [29]

Answer:

Step-by-step explanation: 2/8 =h/2

3 0
2 years ago
Help please due soon
KATRIN_1 [288]

Answer:

y = 6x

Step-by-step explanation:

You can easily find the slope by using slope formula:

(y2-y1) / (x2-x1) (the numbers are subscript)

E.g: (24-12) / (4-2) = 12/2 = 6

m(slope) = 6

Hope this helped :)

(You can also check if it’s correct by inputting the coords into the equation, seeing if all solutions are aplicable).

5 0
2 years ago
Element X decays radioactively with a half life of 6 minutes. If there are 790 grams of Element X, how long, to the nearest tent
REY [17]

Answer:

Step-by-step explanation:idk

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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