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what about questions 2 and 3..?

Let  P  be the price  of  the other  60 lbs  of  nuts

So we have this

40 (.75)  + 60(P)  = 100 (.66)    simplify

30  + 60P  =  66      subtract  30  from both sides

 60P =  36      divide both sides by 60

P =  36 / 60   =    6/10   =   .60    [ 60 cents ]

thank you!!!

These  are binomial  probabilities

(a)  It  fails  for  exactly 6 people  = 

C(40, 6) (.10)^6 (.90)^34   ≈  10.68%

(b)  It fails for at most 2 people  =

P(It fails for 0 people)  +  P(It fails for 1 person)  +  P (It  fails for 2 people) =

(.90)^40  +  C(40,1) (.10)(.90)^39  + C(40,2)(.10)^2(.90)^38 ≈  22.29%

(d)  It  fails  for  8 people at most  =

1  - [  P(0)   + P(1)  + P(2)  + P(3)  + P(4)  + P(5)  + P(6) + P(7)  ]  =

1 -   [(.90)^40  +  C(40,1) (.10)(.90)^39  + C(40,2)(.10)^2(.90)^38  +  C(40,3) (.10)^3(.90)^37  + C(40,4)(.10)^4(.90)^36   +  C(40,5) (.10)^5(.90)^35  + C(40,6)(.10)^6(.90)^34  + C(40,7)(.10)^7(.90)^33 ]  ≈

1  -  [ .2229  + .7353  ] ≈   4.18%

(-π/2,π/2) [-π/2,π/2] (0,π) [0,π] (0,2π) [0,2π]

[1, +inf)                 I think.....all of the other intervals listed would have a cos x  value  of zero included which would result in a zero denominator.

Nice, EP  !!!!

If AD  is parallel  to  BE....then  angles   ADC  and BEC  are equal

But  angle   ACD  is  greater than angle  BCE

So.. in triangles  CAD  and CBE.......this means that  angle  CAD  is less than angle  CBE

cos x         give it amplitude 4

4 cos x      3 units down makes it     4 cos x    -  3

NOT reflected , so it is still   4 cos x    -3

period = 3pi/4    makes it   2pi/(3pi/4)  = 8/3       4 cos (8x/3)   - 3

The directrx  is  below the focus.....so this parabola turns upward

p =  the  distance  between  the  focus and directrix  / 2   =   (4 - - 6)  / 2  = 10 /2  = 5

So the  vertex =   ( - 4,  4 - p)   =  ( -4, 4 -5)  = (-4, -1)

So  we  have   this form

4p ( y - k)  =  (x - h)^2            where  (h, k)  is the vertex.......so....we have

4(5) ( y -  -1)   = ( x   + 4)^2      simplify

20 (y + 1)  = ( x + 4)^2

20y + 20   = x^2  + 8x  + 16       subtract  20 from both sides

20y =   x^2  + 8x   - 4      divide  both sides by 20

y = (1/20)x^2 + (2/5)x  - (1/5)

Here's the graph  :  https://www.desmos.com/calculator/5tzekvcuhj

Looks incorrect to me....it should be a normal sine curve shifted down three units.....and the period should be 2 pi

    yours has a period of more than 2pi

NVM....

It says (2,3) is wrong, although what you did makes sense to me

1    at x = 5   f(x) = -2      

      at x = 9   f(x) = 14        from 5 to 9 is 4     from   -2 to 14  is  16     16/4 = 4 = average

That's right, you can compute the probabilities, and add them up!  i.e. Anty gets to 0 after 1 steps, 2 steps, 3 steps, and so on.  This gives us e_1 = 1/2*1 + 1/4*2 + 1/8*3 + 1/16*4 + ... By arithmetico-geometric series, e_1 = 2.  Similarly, e_2 = 1/2*2 + 1/4*3 + 1/8*4 + 1/16*5 + ... = 3, so (e_1,e_2) = (2,3).

Car 1 distance = 42 m/h  x 3.5 hr = 147 m

Car 2                = 50 m/h x 3 hr = 150 m

Now use pythagorean theorem     d = sqrt ( 147^2 + 150^2) = 210 miles

What about   f <= 11 1/4    lb

Perhaps you are assuming the weights have to be integers.  The question as stated doesn't specify this restriction.

I don't get it then. 11 works because 11*3 + 3*2 = 39. But 12 doesn't work because 12*3 + 2*2 =40. But 2*2+5 doesn't equal 12. Maybe 10 if the equation f <= 2S +5 is actually f < 2S +5. Is this your aops homework? Because I am doing geometry on aops, lol

Let's first just substitute the sack of flour from the second equation into the first equation. You now get 8S+10 < =40, then 8S <= 30. Then, you get

S = 3 because S = 4 won't work. If S = 3, you get 3F <= 34. One sack of flour is equal to 11.

Let  c  = number of chickens     and  r  =  number of rabbits.

Number of heads equation:   c +   r  =  30     --->   multiply by 4   --->    4c + 4r  =  120 

Number of legs equation:    2c + 4r  =  76     --->   multiply by -1  --->   -2c - 4r  =   -76

Add down the columns:                                                                           2c        =     44

                                                                                                                         c  =  22

                                                                                                                         r  =  8

6x + 3y  >  9

        3y  >  -6x + 9               (subtract 6x)

          y  >  -2x + 3               (divide by 3)

Multiplying out and simplifying the left hand side:  (a + 1/b)(b + 1/c)(c + 1/a)

gives you:  abc + (a + b + c) + (1/a + 1/b + 1/c) + 1/(abc)

Multiplying out and simplifying the right hand side:  (1 + 1/a)(1 + 1/b)(1 + 1/c)

gives you:  1 + (1/a + 1/b + 1/c) + ( 1/(ab) + 1/(ac) + 1/(bc) ) + 1/(abc)

Setting these two equal to each other and subtracting both (1/a + 1/b + 1/c) and 1/(abc) from both sides,

leaves you with:  abc + (a + b + c)  =  1 + 1/(ab) + 1/(ac) + 1/(bc)

Rewrite 1/(ab) + 1/(ac) + 1/(bc)  as  (a + b + c)/(abc)

the equation becomes:   abc + (a + b + c)  =  1 + (a + b + c)/(abc)

Since  abc = 13:               13 + (a + b + c)  =  1 + (a + b + c)/13

Subtract 1:                        12 + (a + b + c)  =  (a + b + c)/13

Multiply by 13:            156 + 13(a + b + c)  =  (a + b + c)

Simplify:                     156 + 12(a + b + c)  =  0

                                             12(a + b + c)  =  -156

                                                   a + b + c  =  -13

As follows:

(That last line above should say "There are two prfect squares in that list that satisfy the conditions, ..."

I'll leave you to sum the values!

Find the number of ways of distributing 20 chocolates, 20 gummy bears, and 20 lemon drops among two people, so that each person gets 30 candies. (The candies of each type are indistinguishable.)

If the computer program is properly written and that is the answer then I accept that.

However I am going to attempt if for myself.

Let one of the halves be

x chocolates, y gummy bears and 30-x-y  lemon drops
\(0\le x \le20\\ 0\le y\le20\\ 0\le 30-x-y \le20 \)

the number of dots in the whole square is 21*21 = 441

The number of dots in the triangles is 2(1+2+3+...10) = 2(55)=110

441-110 = 331 combinations ..... maybe

I think there is some double counting here.  

Lets see.

What I have done below is pinpointed all the dots that have been double counted.

The point in the middle has been counted four times.

The lines are 

x=y

x=30-x-y

and 

y=30-x-y

There are 11 points on each of the 3 lines which makes  33 overcounts

331-33= 298 combinations altogether.  This is the same answer that guest got with his/her program.  

Here is my take

3 choices of patty     3    x

9 choices of condiment (I included NO condiment) = 9!

3 x 9! =.......