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rodikova [14]
3 years ago
12

Laura found an old deck of cards in her drawer which had only eight face cards: king of hearts (H), king of spades (S), king of

clubs (C), king of diamonds (D), queen of hearts (H), queen of spades (S), queen of clubs (C), and queen of diamonds (D). Laura and her sister separated the kings and queens into a king deck and a queen deck. They created a game where they draw one card from each deck.
The table below shows all the possible outcomes they can draw.

Outcomes Queen(H) Queen(S) Queen(C) Queen(D)
King(H) King(H),
Queen(H) King(H),
Queen(S) King(H),
Queen(C) King(H),
Queen(D)
King(S) King(S),
Queen(H) King(S),
Queen(S) King(S),
Queen(C) King(S),
Queen(D)
King(C) King(C),
Queen(H) King(C),
Queen(S) King(C),
Queen(C) King(C),
Queen(D)
King(D) King(D),
Queen(H) King(D),
Queen(S) King(D),
Queen(C) King(D),
Queen(D)

What is the probability of Laura and her sister drawing a king that is not the king of hearts, and a queen that is not the queen of spades?
Mathematics
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

9/16

Step-by-step explanation:

Add up all of the ones that don't have the letter you are looking for up and you'll get 9 and all together there is 16 therefore it will be 9/16 :)

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Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64
tino4ka555 [31]

a.

x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}

By Fermat's little theorem, we have

x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5

x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7

5 and 7 are both prime, so \varphi(5)=4 and \varphi(7)=6. By Euler's theorem, we get

x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5

x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7

Now we can use the Chinese remainder theorem to solve for x. Start with

x=2\cdot7+5\cdot6

  • Taken mod 5, the second term vanishes and 14\equiv4\pmod5. Multiply by the inverse of 4 mod 5 (4), then by 2.

x=2\cdot7\cdot4\cdot2+5\cdot6

  • Taken mod 7, the first term vanishes and 30\equiv2\pmod7. Multiply by the inverse of 2 mod 7 (4), then by 6.

x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6

\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}

b.

x^5\equiv3\pmod{64}

We have \varphi(64)=32, so by Euler's theorem,

x^{32}\equiv1\pmod{64}

Now, raising both sides of the original congruence to the power of 6 gives

x^{30}\equiv3^6\equiv729\equiv25\pmod{64}

Then multiplying both sides by x^2 gives

x^{32}\equiv25x^2\equiv1\pmod{64}

so that x^2 is the inverse of 25 mod 64. To find this inverse, solve for y in 25y\equiv1\pmod{64}. Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that (-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}.

So we know

25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}

Squaring both sides of this gives

x^4\equiv1681\equiv17\pmod{64}

and multiplying both sides by x tells us

x^5\equiv17x\equiv3\pmod{64}

Use the Euclidean algorithm to solve for x.

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that (-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}, and so x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}

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Allisa [31]

Answer:

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Step-by-step explanation:

13.8-1=12.8.... unless I'm wrong?

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Answer:

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Step-by-step explanation:

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True [87]
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You need to change 9% into a decimal so it would be 0.09

Then divide 27 ÷ 0.09 = 300

So 9% of 300 is 27
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Answer:

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Step-by-step explanation:

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