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The answer is 15.

The remainder is 2.

I think if x is [0, inf)

B = 1 and C = 3, so B + C = 4.

Set y = 56, so  56 = -16*t^2 + 60*t

Divide by 16 and rearrange this:   t^2 - (15/4)*t + (7/2) = 0

This factorises as (t - 7/4)*(t - 2) = 0

So, there are two times.  Which do you think represents the answer to the question?

Here is my approach:

The answer is 2/5

Think of the cards as all being different 3 different red and 3 different blue

There are 6!  ways that these can be ordered.

now I am going to give the first person 2 cards then the second and then the 3rd.

6 choices to start

then 3 of the ones left are the other colour

now there are 4 cards the second person can get any of them

there are 2 that are a different colour for the seconds person second choice

there are 2 cards left for the 3rd peron to chose from and then just 1

so that is

6*3*4*2*2

so the prob will be  (6*3*4*2*2)/ 6!  =  2/5

you are welcome!

A stick is broken at two points, at random.  If the length of the stick is 6, then what is the probability that all three resulting pieces are shorter than 3 units?

I'd use contour probability mapping

Let one piece be x another  y-x and the third   6-y

we know

0

y-x>0     which simplifies to    y>x

y-x<6      which simplifies to    y

and

0<6-y<6   which simplifies to  -6<-y< 0   or better still    0

Plot all those lines on a number plane.

The area in the middle conatins all the possible outcomes.  

Now you want the section inside where  

x<3

y-x<3  and

6-y<3

Find the area contained in this smaller region

Put the little area over the big area and you have your probability.

The prob that it appears 9 times or more is 1

We know it appears at least 9 times, so the prob that the 'other' roll is also a 6 is  1/6

so I think the answer is  5/6

If the problem is asking for exactly three:

There are 7*4 = 28 total cards
choose 3 colors -> 4c3

choose a number for the three colors to be -> 7c1

choose random other numbers to fill the rest of the hand -> 24c2

(4c3*7c1*24c2)/(28c5)

If asking for three or more: (basically > 3 cards that have the same number)

((ways of getting 3 card same) + (ways of getting 4 cards same)) / (ways to draw 5 cards)

= ((4c3*7c1*24c2)  + (4c4*7c1*24c1))/(24c5)

You should solve it by finding the horizontal asymptope once you graphed the function. Don't worry, it should be pretty simple. 

Ohhhh so you had to use the half-angle identity equation-- thank you so much haha!! I'll try and attempt the question again knowing the formula... thank you so much once again!! Also I understood the concepts on the 2nd and 3rd question on where I went wrong and how to do the question it self. this is really helpful!!

Hmm don't overthink it too much

just add the equations up! You will notice something

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The probability of all three getting a red card is:

[3 / 6]  x  [2 / 5]  x  [1 / 4] ==6 / 120 ==1 / 20

 The sum of all interior angles of a trapezoid==360 degrees

260 - 140 ==220 - sum of the remaining 3 angles.

a+b+c+140 ==360,   

140 - c==c  -  b,   

c  -  b==b  -  a, solve for a, b, c

a ==40 degrees

b ==220 / 3 ==73 + 1/3 degrees

c ==320 / 3 ==106  + 2/3 degrees.

So, the 4 angles are: 40,  73 1/3,  106 2/3,  140

And the smallest angle ==40 degrees.

Without a minimum number of packets to each teacher, then you have:

(12, 2, 1) , (10, 4, 1) , (8, 6, 1) , (6, 8, 1) , (4, 10, 1) , (2, 12, 1) , (10, 2, 3) , (8, 4, 3) , (6, 6, 3) , (4, 8, 3) , (2, 10, 3) , (8, 2, 5) , (6, 4, 5) , (4, 6, 5) , (2, 8, 5) , (6, 2, 7) , (4, 4, 7) , (2, 6, 7) , (4, 2, 9) , (2, 4, 9) , (2, 2, 11) , (12, 1, 2) , (10, 3, 2) , (8, 5, 2) , (6, 7, 2) , (4, 9, 2) , (2, 11, 2) , (10, 1, 4) , (8, 3, 4) , (6, 5, 4) , (4, 7, 4) , (2, 9, 4) , (8, 1, 6) , (6, 3, 6) , (4, 5, 6) , (2, 7, 6) , (6, 1, 8) , (4, 3, 8) , (2, 5, 8) , (4, 1, 10) , (2, 3, 10) , (2, 1, 12) , (11, 2, 2) , (9, 4, 2) , (7, 6, 2) , (5, 8, 2) , (3, 10, 2) , (1, 12, 2) , (9, 2, 4) , (7, 4, 4) , (5, 6, 4) , (3, 8, 4) , (1, 10, 4) , (7, 2, 6) , (5, 4, 6) , (3, 6, 6) , (1, 8, 6) , (5, 2, 8) , (3, 4, 8) , (1, 6, 8) , (3, 2, 10) , (1, 4, 10) , (1, 2, 12)==63 different ways

Note: The sum of each permutation ==15 and each permutation consists of: Two EVEN numbers + one ODD number.