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có 11 chiếc thẻ được đánh số từ 1 tới 11, người ta rút ngẫu nhiên 2 thẻ khác nhau. Xác suất để rút được 2 thẻ mà tích 2 số được đánh dấu trên thẻ là số chẵn ?

google translates that as: There are 11 cards numbered from 1 to 11, they draw randomly 2 different cards. What is the probability of drawing 2 cards with a 2-digit tick on the card?    

Since there are 11 cards and they're numbered 1 to 11,

then there is only one card in the deck that has a 2 on it. 

Therefore, you can't draw two cards with a 2 on them.  =  Probability zero 

_.

See https://web2.0calc.com/questions/help_21753

When subtracting two different numbers from that set, you can get every positive answer from 1 through 15.

Solve:     (2/3)·(a - 4)  =  -3(a - 3)

Multiply both sides by 3:            3·[ (2/3)·(a - 4) ]  =  3·[ -3·(a - 3) ]

                                                               2·(a - 4)  =  -9·(a - 3)

Distribute:                                                   2a - 8  =  -9a + 27

Add  9a  to both sides;                              11a - 8  =  27

Add  8 to both sides:                                      11a  =  35

Divide both sides by 11:                                   a  =  35/11

To get one and only one of the two counters to be white, you can either

1)  get a white counter followed by a non-white counter:  (6/10)·(4/9)  =  24/90

2)  or get a non-white counter followed by a white counter:  (4/10)·(6/9)  =  24/90

To find the final probability, add those two probabilities together and reduce your answer.

First one, just do the multiplication   (1/3)(30 units²)(6.5 units)  =  65 units³  

Second one, just add up like terms:  

                   +3x²     –2x     +1    

                   –6x²     +5x     –4    

                   +9x²     –8x     +7  

                  ——–   ––––   –––    

Totals          +6x²     –5x     +4    so the answer is 6x² – 5x + 4   

_.

94%  =  94 / 100  =  47 / 50

So, for the winning percentage to be exactly 94%, Teddy must have played at least 50 games.

What is the smallest integer n, greater than 1, such that \(n^{-1}\pmod{130}\) and \(n^{-1}\pmod{231}\) are both defined?

Modular multiplicative inverse is defined if \(\gcd(130,n)=\gcd(231,n)=1\)

\(\begin{array}{|rcll|} \hline \text{factor}~ 130 &=& 2*5*13 \\ \text{factor}~ 231 &=& 3*7*11 \\ \hline \end{array}\)

The first prim number = 2. This number is a prime factor in 130, so \(\gcd(130,2) \ne 1\)
The 2nd prim number = 3. This number is a prime factor in 231, so \(\gcd(231,3) \ne 1\)
The 3rd prim number = 5. This number is a prime factor in 130, so \(\gcd(130,5) \ne 1\)
The 4th prim number = 7. This number is a prime factor in 231, so \(\gcd(231,7) \ne 1\)
The 5th prim number = 11. This number is a prime factor in 231, so \(\gcd(231,11) \ne 1 \)
The 6th prim number = 13. This number is a prime factor in 130, so \(\gcd(130,13) \ne 1\)
The 7th prim number = 17. This number is not a prime factor in 130 and not a prime factor in 231
, \(\gcd(130,17) = \gcd(213,17) = 1\)

The smallest integer n is 17

check:

\(\begin{array}{|rcll|} \hline 17^{-1} \pmod{130} &\equiv& 23 \\ 17*23 & \equiv& 1 \pmod{130}~ \checkmark \\\\ 17^{-1} \pmod{231} &\equiv& 68 \\ 17*68 & \equiv& 1 \pmod{231}~ \checkmark \\ \hline \end{array}\)

How about 53C3   =    23426   

the line is dashed which means it is not included so there cannot be an = as a part of the answer.

Just test a point that is not on the line (0,0) is the easiest one, and see which equation makes this true.

I mean sub x=0 and y=0 into the 2 remaining possible answers and only one of them will be true.

Next time please present your question properly by deleting the dollar signs

\(\lfloor x \rfloor + x = \dfrac{13}{3}\\ \lfloor x \rfloor + x =4 \dfrac{1}{3}\\ \text{obviously}\;\;\lfloor x \rfloor\;\;\text{is a whole number }\\ so\\ \lfloor x \rfloor + \lfloor x \rfloor + \dfrac{1}{3} = 4 \dfrac{1}{3}\\ so\;\;\;\lfloor x\rfloor=2\\ x=2\frac{1}{3} \)

I was referring  to polyyou's assumption, which was based on no fact.

You have a triangle with 45 identical lines in this way as you can see on the picture.

What is gamma?

\(\begin{array}{|r|r|} \hline & angle \\ \text{lines} & =\dfrac{\text{lines}-1}{2} \\ \hline 5 & 2\gamma \\ 7 & 3\gamma \\ 9 & 4\gamma \\ 11 & 5\gamma \\ 13 & 6\gamma \\ 15 & 7\gamma \\ \ldots & \ldots \\ 45 & \dfrac{\text{45}-1}{2}=22\gamma\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2*22\gamma+\gamma &=& 180^\circ \\ 45\gamma &=& 180^\circ \\ \gamma &=& \dfrac{180^\circ}{45} \\ \mathbf{\gamma} &=& \mathbf{ 4} \\ \hline \end{array}\)

If you mean to do it in your head, then multiply 3456 by 100 and then subtract 3456.

OK then, with all the following restrictions:

1 - That all 5 digits be odd

2 - That they all be distinct or different from each other.

3 - That  no two adjacent digits can be more than 6 apart [I took this to mean that the difference between any two adjacent digits be 6 or less)

4 - If I got all the conditions right, then you have the following numbers:

13579 , 13597 , 13759 , 13795 , 13957 , 13975 , 15379 , 15397 , 15739 , 15793 , 15937 , 15973 , 17359 , 17395 , 17539 , 17593 , 17935 , 17953 , 31579 , 31597 , 31759 , 31795 , 35179 , 35971 , 37159 , 37951 , 39517 , 39571 , 39715 , 39751 , 51379 , 51397 , 51739 , 51793 , 53179 , 53971 , 57139 , 57931 , 59317 , 59371 , 59713 , 59731 , 71359 , 71395 , 71539 , 71593 , 73159 , 73951 , 75139 , 75931 , 79315 , 79351 , 79513 , 79531 , 93157 , 93175 , 93517 , 93571 , 93715 , 93751 , 95137 , 95173 , 95317 , 95371 , 95713 , 95731 , 97135 , 97153 , 97315 , 97351 , 97513 , 97531 , Total =  72

I thought i already put this here, but anyway...

Thank you, Dragan!

Thank you, Dragan !