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Kryger [21]
3 years ago
7

Which relationship between x and y in the equation shows a proportional relationship?

Mathematics
2 answers:
amid [387]3 years ago
4 0

Its c. This is for the other guys brainliest:)

LUCKY_DIMON [66]3 years ago
3 0

its C i just commented this so you can mark the other guy brainliest

:3

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Five cards are dealt from a standard 52-card deck. (a) What is the probability that we draw 1 ace, 1 two, 1 three, 1 four, and 1
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Answer:

Step-by-step explanation:

As there are total 52 cards in a deck and we have to draw a set of 5 cards, we can use the formula of combination to find the total number of possible ways of drawing 5 cards.

Number of ways to draw 5 cards = N_T

N_T\;=\;({}^NC_k)\\\\N_T\;=\;({}^{52}C_5)\\\\N_T\;=\;2,598,960

(a) Assuming the cards are drawn in order (would not affect the probability). The of getting Ace, 2, 3, 4 and 5 can be obtained by multiplying the probability of getting cards below 6 (20/52) with the probability of getting 5 different cards (4 choices for each card).

P(a)\;=\;\frac{20}{52}*\frac{4}{52}*\frac{4}{51}*\frac{4}{50}*\frac{4}{49}*\frac{4}{48}\\\\P(a)\;=\; 1.3133*10^{-6}

(b) For a straight we require our set to be in a sequence. The choices for lowest value card to produce a sequence are ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. Hence, the number of ways are ({}^{10}C_1).

For each card we can draw from any of the 4 sets. It can be described mathematically as: ({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)*({}^{4}C_1)\;=\;[({}^{4}C_1)^5]

Therefore, the total outcomes for drawing straight are:

N_S\;=\;({}^{4}C_1)*({}^{4}C_1)^5\;=\;10240

Thus, the probability of getting a straight hand is:

P(b)\;=\;\frac{N_S}{N_T}\\\\P(b)\;=\;\frac{10240}{2598960}\\\\P(b)\;=\; 0.0039

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