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We will  have an equation of this form

Height  = A cos ( Bx  + C)  + D

A  =  the amplitude  =   [ 14 - 4 ] / 2  =  5

The period =  10 hrs....so   10  =  2pi / B   ⇒ B  = 2pi/10 =  pi/5

D  is the  midline   =  [ 14 + 4 ] /2  =  18 /2   = 9

C  is the phase shift  and is a little tough to figure

At 2AM,   x  = 2,  the height is  4 ft.....so we need to solve this

4  =  5 cos ( pi/5 * 2 +  C )  +  9     subtract  9 from both sides

-5  = 5 cos ( 2pi/5 + C)     divide both sides  by  5

-1  = cos  (2pi/5 + C)         we  need to take the  arccos

arccos (-1)  =  2pi/5 +C

pi  = 2pi/5  + C

C  =  pi - 2pi/5 = [  5pi  -2pi] / 5  =  3pi/5  

So....the function is

Height =  5cos [ ( pi /5) x    + 3pi/5 )  +  9

Here's the graph  : https://www.desmos.com/calculator/igberusuwx

The number 15 has four factors (1, 3, 5, 15), and the number 9 has three factors (1, 3, 9), so the total number of possible rational roots is 4*3 = 12.

Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm pretty good at it: If I shoot 5 pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability 211/243. If I shoot 6 pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin?

Edit:

NOTE: I have given the correct answer to the WRONG question.  There is another post below where I have answered the question that was actually asked    

If I shoot 5 bits the prob that they all miss is   

\( P(all\;fail\;on\; 5\; tries)=5C0*(F)^5=F^5\\ P(don't \;all\;fail\;with\; 5\; tries)=1-F^5=\frac{211}{243}\\ so\\ F^5=\frac{243-211}{243}=\frac{32}{243}\\~\\ F=\frac{2}{3}\qquad S=\frac{1}{3} \)

\( P(none\; in)=6C0\;* \; (\frac{1}{3})^0\;*\;(\frac{2}{3})^6\\ P(none\; in)=1\;* \; 1\;*\;(\frac{2}{3})^6\\ P(none\; in)=(\frac{2}{3})^6\\ P(none\; in)=\frac{64}{729} \\~\\ P(one\; in)=6C1\;* \; (\frac{1}{3})^1\;*\;(\frac{2}{3})^5\\ P(one\; in)=6\;* \; (\frac{1}{3})\;*\;(\frac{32}{243})\\ P(one\; in)=2\;*\;(\frac{32}{243})\\ P(one\; in)=\frac{64}{243}\\ \\~\\ P(two\; in)=6C2\;* \; (\frac{1}{3})^2\;*\;(\frac{2}{3})^4\\ P(two\; in)=15\;* \; (\frac{1}{9})\;*\;(\frac{16}{81})\\ P(two\; in)=5\;* \; (\frac{1}{3})\;*\;(\frac{16}{81})\\ P(two\; in)= \frac{80}{243}\\ \\~\\ P(\text{no more than 2 in)}=\frac{64}{729}+\frac{64}{243}+ \frac{80}{243}\\~\\ P(\text{no more than 2 in)}=\frac{64+3*64+3*80}{729}\\ P(\text{no more than 2 in)}=\frac{496}{729}\\ \)

\(P(\text{no more than 2 in)}\approx 0.68\)

There could easily be careless mistakes, maybe even careless logic mistakes, but the underlying idea is sound.

My answer almost agrees with the second guest, and much of the logic is in agreement with Gino.

So we all had similar ideas.

Coding:
P(all\;fail\;on\; 5\; tries)=5C0*(F)^5=F^5\\
P(don't \;all\;fail\;with\; 5\; tries)=1-F^5=\frac{211}{243}\\
so\\
F^5=\frac{243-211}{243}=\frac{32}{243}\\~\\
F=\frac{2}{3}\qquad S=\frac{1}{3}

P(none\; in)=6C0\;* \; (\frac{1}{3})^0\;*\;(\frac{2}{3})^6\\
P(none\; in)=1\;* \; 1\;*\;(\frac{2}{3})^6\\
P(none\; in)=(\frac{2}{3})^6\\
P(none\; in)=\frac{64}{729}
\\~\\
P(one\; in)=6C1\;* \; (\frac{1}{3})^1\;*\;(\frac{2}{3})^5\\
P(one\; in)=6\;* \; (\frac{1}{3})\;*\;(\frac{32}{243})\\
P(one\; in)=2\;*\;(\frac{32}{243})\\
P(one\; in)=\frac{64}{243}\\
\\~\\
P(two\; in)=6C2\;* \; (\frac{1}{3})^2\;*\;(\frac{2}{3})^4\\
P(two\; in)=15\;* \; (\frac{1}{9})\;*\;(\frac{16}{81})\\
P(two\; in)=5\;* \; (\frac{1}{3})\;*\;(\frac{16}{81})\\
P(two\; in)= \frac{80}{243}\\

\\~\\
P(\text{no more than 2 in)}=\frac{64}{729}+\frac{64}{243}+ \frac{80}{243}\\~\\
P(\text{no more than 2 in)}=\frac{64+3*64+3*80}{729}\\
P(\text{no more than 2 in)}=\frac{496}{729}\\

A family buys a car for $20,000. The value of the car decreases about 20% per year. After 6 years, the family decides to sell the car. Should they sell it for $4,000? Explain

Car value today is $ 20,000

year

1st         20,000    -20%   = 16,000

2nd        16,000    -20%   = 12,800

3rd         12,800    -20%   = 10,240

4th          10,240   -20%   =  8,192

5th          8,192     -20%   = 6,553.6

6th          6,553.6   -20%   = 5,242.88

445  employees take a train to work  and  321  take only a triain

So

445 -  321 =   124   take both a bus and train to work

So.....621 - 124  =  497  take only a bus 

So  the probability  that an employee  takes  a  bus or train (but not both)  =

[ 321  + 497 ]  /  1108   =

818 / 1108   ≈  .738  =   73.8%

Let  the number of hours that  Rosa takes to write her essay  =  x

So....each hour.....she completes   1/x   of the essay

And the number  of hours  that Leann takes to write her essay  =  x + 6

So....each hour.....she completes  1 / [ x + 6]   of the essay

So....working together  they take  4  hours to complete the essay.....which means that each hour working together, they complete (1/4)  of the essay

So  we have that

1/x  +  1 / [ x + 6]  =  1/4    

[x + 6 + x ]            1

__________  =   ___

 x (x + 6)               4

2x + 6               1

________  =   ____              cross-multiply

x^2 + 6x            4

4 ( 2x + 6)  =  x^2 + 6x

8x + 24  = x^2  + 6x     rearrange as

x^2  - 2x  - 24   =   0      factor

(x - 6) ( x + 4)  =  0

Setting the first factor to 0  and solving  for  x    gives that  x  = 6

So...it takes Leann   x + 6 =   6 + 6   = 12  hrs to write the essay by herself 

The vertex  is  (-3, 5)

And the x coordinate of the  vertex  is given by

-b / [ 2a]  = -3

b = 6a

And the point  (-1, 3)  is on the graph.....so.....we have this system

3  =  a(-1)^2  + 6a (-1)  + c

5  =  a(-3)^2  + 6a(-3)  + c         simplify

a  - 6a  + c   = 3

9a  - 18a  + c  = 5

-5a  + c   = 3

-9a  + c   = 5          multiply  the second equation through by  -1

-5a  + c   = 3

9a   -  c  = -5           add these

4a  = -2

a = -2/4  = -1/2  

So  b  = 6(-1/2)  =  -3

And  -9a  + c  =  5    ....so....

-9(-1/2) + c  = 5

9/2  + c  =10/2

c = 10/2  - 9/2 =   1/2

So

a  *  b  *  c  =

(-1/2) * (-3)  * 1/2  =

3 / 4

Here's a graph  : https://www.desmos.com/calculator/nl6jcccb3k

The red parabola shown is the graph of the equation $y=ax^2 + bx + c$. Find $a\cdot b\cdot c$.

617 was correct! Thank you both and have a good day :D

The parabola is -1/3*x^2 - 2x + 2, so abc = (-1/3)(-2)(2) = 4/3.

To get a matching pair, he will need either 2 blue socks or 2 red socks or 2 white socks or 2 green socks.

There are (5 + 7 + 3 + 4) = 19 socks.

To get 2 blue socks, he will need a blue sock (5/19) and another blue sock (4/18)        =  (5/19)(4/18)  =  20/342.

To get 2 red socks, he will need a red sock (7/19) and another red sock (6/18)             =  (7/19)(6/18)  =  42/342.

To get 2 white socks, he will need a white sock (3/19) and another white sock (2/18)    =  (3/19)(2/18)  =  6/342.

To get 2 green socks, he will need a green sock (4/19) and another green sock (3/18)  =  (4/19)(3/18)  =  12/2342.

Any of those combinations will give him a matching pair of socks, so add them:  

     20/342 + 42/342 + 6/342 + 12/342  =  80/342.

This is similar  to  distributing k identical balls into n distinct boxes......where  we have no restrictions [ any box can be empty ]  

This is given by

C ( n + k  -1 , n -1)   =  C ( 3 + 6 - 1 , 3 - 1)  =  C ( 8, 2)   =   28 ways

In right triangle ABC with the right angle at vertex A, altitude AE is drawn from vertex A to hypotenuse BC, AE = 21, and CE = 20. Find the length of BC. Round your answer to the nearest hundredth.

Triangles  ABC, ACE and ABE  are similar.

AE = 21

CE = 20

Angle ACB = ?              tan(ACB) = AE / CE        ∠ ACB = 46.397°

BE = tan(BAE) * AE           BE = 22.05

BC = CE + BE           BC = 42.05  

Yeah....I knew there was a formula but I forgot exactly what it was......!!!

The probability  is

(0.85)^3   ≈ .614  =    61.4 %

Impressive!!!  

Ohhh.

Well, we can multiply 40 green cars by 3 to even the ratio out so it's 3 to 3.

So that means there are 120 cars in order to be 3 to 3, which means 120 silver cars.

Subtracted 20 from 120 = 100  because 20 silver cars. So that means there are 100 silver cars.

Note that  BC  = (1/3)AB

We have that  C  = 

[ -1  +  (1/3) ( -1  - - 7)  ,  0  + (1/3)( 0 - 3)  ]

[ -1 + (1/3)(6)  ,  0 + ( 1/3)(-3)  ]

[ -1 + 2 , 0 - 1]

( 1, - 1)

Depends on how many you HAD before you stole 2 more from the counter !  

Thanks, CoolStuffYT.....!!!!

Here's one more way  :

Note that  triangles  CAE  and CBA are similar

And using CoolSuffYT's   answer that  AC  =29   we have that

So

AC / EC  =  BC  / AC

29 / 20  =  BC  /  29

29^2 / 20  =  BC

841 / 20   = BC

42.05  =  BC

The question is how many do YOU have, not how many are left.
So two, since you took two.

Let the rate of the current  =  R

Note that  15 min   = 1/4  hr

So  we have that 

The time with the current  +  1/4 hr  =  time against the current  .....and with the current the rate is 15 + R  .....and against the current the rate is  15 - R ......and   distance / rate  = time

So

   5                1                5

______  +   ____  =   ______     rearrange  as

15 + R           4           15  - R

   5               5                      1

______  - ______  =     -   ____

15 + R      15 - R                 4

5 ( 15 - R) - 5(15 + R)                  1

_________________  =   -    ______

(15 + R) ( 15 - R)                         4

  -10R                     -1

________  =       ______      cross-multiply

225 - R^2                4

-40R  =  225 - R^2

R^2 - 40R - 225  =  0       factor

(R - 5)  ( R + 45)   = 0

Only the first factor set to  0  gives us the speed of the current

R -  5   =   0

R  = 5  (mph)

With pascal's triangel:

Isn't it 0? 2 ducks, take away two ducks. 0 ducks?