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Juli2301 [7.4K]
3 years ago
10

PLEASE HELP ITS DUE TMR

Mathematics
2 answers:
maks197457 [2]3 years ago
7 0

Answer:

Choice C 42 in^2

Step-by-step explanation:

Area of a triangle= 1/2(b)(h)

= 1/2(3)(6)

= 9

Since there is two triangles we can do:

9(2) = 18

Area of a rectangle = (b)(h)

= (4)(6)

= 24

24+18 = 42 in^2

sattari [20]3 years ago
6 0

Answer:

it's 42 my bad...

Step-by-step explanation:

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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.

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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]

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

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

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