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its bison

Bison

Mary has six cards whose front sides show the numbers  1,2,3,4,5and 6 .
She turns the cards face-down, shuffles the cards until their order is random,
then pulls the top two cards off the deck.
What is the probability that at least one of those two cards shows a square number?

\(\color{red}\text{square number}\)

\(\begin{array}{|r|r|r|r|r|r|r|r|} \hline & \text{card 2} & {\color{red}1} & 2 & 3 & {\color{red}4} & 5 & 6 \\ \hline \text{card 1} & & & & & & & \\ \hline {\color{red}1} & & - & \times & \times & \times & \times & \times \\ \hline 2 & & \times & - & & \times & & \\ \hline 3 & & \times & & - & \times & & \\ \hline {\color{red}4} & & \times & \times & \times & - & \times & \times \\ \hline 5 & & \times & & & \times & - & \\ \hline 6 & & \times & & & \times & & - \\ \hline \end{array}\)

\(\text{The probability is $\dfrac{18}{30} = \dfrac{3}{5} \quad (60\ \%) $ } \)

This answer is incorrect.

heart pls

heart pls

Piecewise Graphs:

\(f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases} \)

f(x) = (backslash)     begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}          

I did not put that latex in the latex box. It was supposed to just display as code.

The latex just keeps showing me things on this post!

\(f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}\)

The LaTex for this question is really cool.  I've not seen this code before.

f(x) = \begin{cases} k(x) &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases}

----------------------------------------

\(y=x^2-6x+12\\ y=x^2-6x+9+3\\ y=(x-3)^2+3\\ y-3=(x-3)^2\\ \pm \sqrt{y-3}=x-3 \qquad where\;\;y\ge3\\ x=\pm \sqrt{y-3}+3\\ \text{The desired inverse will be one of these.}\\ y=\pm \sqrt{x-3}+3\\ \text{BUT is it the plus or the minus?}\)

Consider f(0)=12  

The reflection of (0,12) in the line y=x (which is the inverse) is  (12,0)

This works when the sign is negative.

so this underneath will be its own inverse.

\(f(x) = \begin{cases}3 -\sqrt{x-3} &\text{if }x>3, \\ x^2-6x+12&\text{if }x\leq3. \end{cases} \)

And here is the graph:

Meyer rolls two fair, ordinary dice with the numbers  on their sides.
What is the probability that at least one of the dice shows a square number?

\(\color{red}\text{square number} \)

\(\begin{array}{|r|r|r|r|r|r|r|r|} \hline & \text{dice 2} & {\color{red}1} & 2 & 3 & {\color{red}4} & 5 & 6 \\ \hline \text{dice 1} & & & & & & & \\ \hline {\color{red}1} & & \times & \times & \times & \times & \times & \times \\ \hline 2 & & \times & & & \times & & \\ \hline 3 & & \times & & & \times & & \\ \hline {\color{red}4} & & \times & \times & \times & \times & \times & \times \\ \hline 5 & & \times & & & \times & & \\ \hline 6 & & \times & & & \times & & \\ \hline \end{array} \)

\(\text{The probability is $\dfrac{20}{36} = \dfrac{5}{9} \quad (55.6\ \%) $ } \)

The denominator of a fraction is 3 more than its numerator.

When  12   is added to this fraction, the resulting fraction's denominator is twice the denominator of the original fraction and its numerator is 1 more than its denominator.

What is the numerator of the original fraction?

Let the fraction be    \(\frac{a}{a+3}\)

\(\frac{a}{a+3}+12 =\frac{a+12(a+3)}{a+3}=\frac{13a+36}{a+3}=\frac{2(13a+36)}{2(a+3)}\)

The meaning of the question is very unclear.

I've tried a couple of different versions an none make sense.

This question,as far as I can tell, is a garbage question.

You do not need to know which angles are which and please, note that this figure is not drawn to scale. You have to first calculate the sum of interior angles and see now many angles degrees are missing. Then see which of the interior angles supplied fit with this sum to make it 360.

\(\text{The only two numbers appearing on each die that are perfect squares are }1 \text{ and } 4\\ P[\text{1 or 4 appear on at least 1 die out of 2}] = 1 - P[\text{no 1s or 4s}] = \\ 1-\dfrac{4^2}{36} = \dfrac{20}{36} = \dfrac{5}{9}\)

A collection of nickels, dimes, and quarters is worth 5.30. There are two more dimes than nickels and four more quarters than dimes. How many quarters were in the collection?

this is an international forum. You should spell out the value of your currency.

I think...

You mean 

$5.30

a nickel is 5c 

a dime is 10c

a quarter is 25c

Let N be the number of nickels, D be the number of dimes and Q be the number of quarters.

There are two more dimes than nickels

D=N+2

and four more quarters than dimes.

Q=D+4

so

Q=N+2+4=N+6

-----

5N+10D+25Q=530

N+2D+5Q      = 106

N+2(N+2)+5(N+6)=106

N+2N+4+5N+30=106

8N+34               =  106

             8N        = 72

                     N   =9

so

N=9      D=11   and   Q=15

check  9*5+11*10+15*25 = 530         Das ist ausgezichnet!

So wir haben

9 nickels,  11 dimes, and  15 quarters.

A regular hexagon has vertices at $1-i, 0, a, b, c, d$ clockwise in that order, as in the picture below:

Then what is \(\operatorname{Im}(a)\)?

\(\begin{array}{|rcll|} \hline \sin(75^{\circ}) &=& \dfrac{\operatorname{Im}(a)} {\sqrt{2}} \\ \operatorname{Im}(a) &=& \sqrt{2}\cdot \sin(75^{\circ}) \\ \operatorname{Im}(a) &=& 1.41421356237\cdot 0.96592582629 \\ \mathbf{\operatorname{Im}(a)} & \mathbf{=} & \mathbf{1.36602540378} \\ \hline \end{array}\)

Thank you

Mary has six cards whose front sides show the numbers  1,2,3,4,5and 6 . She turns the cards face-down, shuffles the cards until their order is random, then pulls the top two cards off the deck. What is the probability that at least one of those two cards shows a square number?

The only square numbers are 1 and 4

P(one of them is 4) = \(\frac{1}{6}\times \frac{5}{5}+\frac{5}{6}\times \frac{1}{5}=\frac{2}{6}\)

P(one of them is 1) = \(\frac{2}{6}\)

P(one is 1 and the other is 4)=\(2*\frac{1}{6}\times\frac{1}{5}=\frac{2}{30}\)

so 

P(at least one is either a 1 or  a 4 is) =  \(\frac{4}{6}-\frac{2}{30}=\frac{18}{30}=\frac{3}{5}\)

Happy New Year to you too   

Is 4 the only thing you consider a square number?          Or is six kinda square?    Or is '1' a square (1^2 = 1...so one is a square)...

   need clarification   .....do you think 5 is a square on the die?

Nah...... there are numerous instances where the questioner's numbers get repeated   9090 is really 90    3232 is actually 32....dont know why it happens, but it does fairly regularly. 

Thank you!

EP i think you misread everything in your equation would be multiplied by 101.

so there are 1919 chickens and 1313 horses.

HOPE THIS HELPED!