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3 years ago

Find m ABD. Please help me.

Mathematics

1 answer:

vlada-n [284]3 years ago

3 0

Answer:

31x

Step-by-step explanation:

Were just looking at the top triangle because it has the points A,B, and D. now we solve for x (aka answer the equation in the parentheses). We need to get rid of the 22 so we only have x. We do that by adding it to itself so it equals 0. What we do to one side we do to the other. Now adding 22 to 9 we get 31 (put the x back).

9x - 22

+22 +22

9x - 22 = 31x

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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

labwork [276]

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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ANSWER:
t=9

EXPLANATION:
-t = 9t - 90
90 = 10t
t = 9

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