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P (4 green peppers)  =   C(5,4) * C (10,2)  / C(15 ,6)    = 45  /1001

P(5 green pepers)   = C(5,5) * C(10,1)  /C(15,6)    =  2 /1001

Total probability   =  [45 + 2]  /1001  =   47 /1001

It is quite complicated to consider those numbers which have a digit sum that is at least 6. 

We can consider those numbers which have a digit sum less than 6 instead.

Numbers with digit sum 4: 1111

Numbers with digit sum 5: 2111, 1211, 1121, 1112

Therefore, there are only 5 numbers with a digit sum less than 6. All other numbers have a digit sum of at least 6.

Note that there are 3^4 = 81 numbers which can be written with only 1, 2, 3.

\(\text{Required number} = 3^4 - 5 = 76\).

According to Wolfram Alpha, the only real value that satisfies this equation is \(x = \dfrac13 \left(4 - \dfrac{\sqrt[3]{4}}{\sqrt[3]{113+3\sqrt{1419}}} + \sqrt[3]{2\left(113+3\sqrt{1419}\right)}\right)\).

Factor the left-hand side of the second equation: \(2a(b+5)=3b+257\)

Do the same thing to the other side: \(2a(b+5)=3(b+5)+242\)

Isolate the constant and simplify: \((2a-3)(b+5)=242\)

The only "reachable" factors of 242 are 22 and 11. 

We can achieve these numbers when \(a = 7\) and \(b = 17\).

Thus, \(ab = 17 \times 7 = \color{brown}\boxed{119}\)

Ha!   BB is correct !    I do not know why I typed that last line      Corrected... THANX, BB !

   -b = -2.50

            b = $ 2.50

Let  x  be the number of $12 increases

R =  Tickets sold *  Cost per ticket

R =  ( 40000 - 500x) ( 200 + 12x)

R =  8000000 - 100000x + 480000x - 6000x^2

R  =   -600x^2 + 380000x + 8000000

The  number of  $12 increases  (which is an integer ) that max the revenue =   [  -380000 / [ 2(-6000)] ] = 31 + 2/3    ≈  32

The cost per ticket that maxes the revenue ≈   (200 + 12 *32)  ≈ $584

\(x =\)  The number of times GetThere Airlines increases the ticket cost by \($12\)

Write as an equation: \((200+12x)(40,000-500x)\)

Simplify: \(−6000x^2+380000x+8000000\)

The vertex appears at \(-380000\div -12000 = 31 {2 \over 3} \approx 32\)

Thus, they make the most money when the tickets are priced at: \(200 +32 \times 1 2 = \color{brown}\boxed{584}\)

Graph: https://www.desmos.com/calculator/t08bdptj6k

I forgot to attach the graph when I sent the response.

Edit: Updated the graph to the new one.

This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3.

We can view this geometrically.

Let \(D\) be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5).

Let \(\Omega\) be the region bounded by 0 < x < 3, y - x < 3, 5 - y < 3, x < y and \(D\).

Basically, the probability is equal to \(\dfrac{\text{area}(\Omega)}{\text{area}(D)} = \dfrac{9 - 2 - \dfrac12}{25} = \dfrac{13}{50}\).

I believe that is the answer.

Edit: Edited. This should be correct now.

The answer isn't mine tho... 

It's Melody's and I trust them more than I trust myself...

We can choose  any 4 of the 9   people to be on  1 of the 3 teams

And then we can choose any 4 of the 5 remaining people to be on 1 of the remaining 2 teams

The last person is on the last team by default

So

C ( 9,4)  * C ( 3,1) * C(5,4) * C (2,1)  =   3780  ways

Not totally confident about this, but BuiderBoy  will  probably have the correct answer  !!!!  (LOL !!! )

Note that by property of centroid, AG : GD = 2 : 1.

Since MN is parallel to BC, MG is parallel to BD and GN is parallel to DC.

Since MG is parallel to BD, \(\triangle AMG \sim \triangle ABD\) by AA postulate.

Since GN is parallel to DC, \(\triangle AGN \sim \triangle ADC\) by AA postulate.

Then, \(MG = \dfrac23 BD = \dfrac23 DC = GN\).

Hence, G divides MN internally in the ratio 1 : 1.

Therefore, area of triangle AGN = area of triangle AGM = 144.

There are 9 ways to choose a single person.

You then have \(8 \choose 4\) ways to choose a group of 4 people. But, you have to divide by 2, to correct for overcounting.

Once that is done, you have \({4 \choose 4} = 1\) way to choose the remaining. 

Thus, there are \(9 \times {8 \choose 2} \div 2 = \color{brown}\boxed{315}\) ways.

All credit goes to Melody: https://web2.0calc.com/questions/a-group-of-nine-people-is-be-split-into-three-teams_1

I was away about nine months, too....it was a little slow when I first came  back, but it appears to have "revived" some !!!!

Nice to see you too, CPhill :)

It has been a long time since I was last active on this site. It feels great to come back to this site and see that it is still a very active forum.

Whoopee !!!

Also.....good to see you on here, Max  !!!!!

Nice!

Ayoo... That's my 400th answer

g(0) is not well-defined since f^{-1}(0) does not exist in [0, \infty), i.e., there is no x \in [0, \infty) such that f(x) = 0.

Proof:

Since \(x\in [0, \infty)\), \(x^3, x^2, x\) are all nonnegative.

Then \(f(x) = x^3 + x^2 + x + 2 \geq 0 + 0 + 0 + 2 = 2 > 0\) for any \(x\in [0, \infty)\).

Then we have \(f(x) > 0\) for any \(x\in [0, \infty)\).

This implies there are no \(x\in [0, \infty)\) such that \(f(x) = 0\).

Therefore, we cannot find the value of \(g(0) + g(3)\).

happy birthday melody

Let \(h\) be the original height and suppose a decrease of \(k\%\) is required.

Let \(r\) be the original radius,

Now,

\(\pi r^2 h = \pi (r(1 + 36\%))^2(h(1 - k\%))\\ (1.36)^2 (1 - k\%) = 1\\ k\% = 1 - \dfrac1{1.36^2}\\ k\% = \dfrac{531}{1156}\\ k = \dfrac{13275}{289}\)

Therefore, the height needs to be decreased by \(\dfrac{13275}{289}\%\), which is approximately \(45.9\%\).

Without loss of generality, let the diameter be 100, and let the height be 100. 

The original volume is: \(250000 \pi\)

With the increased base, the volume of the new cube is: \(462400\pi\)

For the volume to stay the same, we have to multiply the height by \({250000 \over 462400} = {625 \over 1156}\)

This means that the percent decrease in the height is \(\approx \color{brown}\boxed{45.93 \text{%}}\)

Let the original volume =  pi * r^2 * original height

Note that if the diameter is increased by 36%, then so is the radius

So

pi * r^2 *  original height      =   pi *  (1.36r) ^2  * new height

original height =   (1.36)^2  * new height

original height / (1.36)^2   =new height

original height  * ( 1 /1.36^2)  =  new height

.54 original height = new height

The original height  should be reduced by  ≈   (1 - .54)  =  .46 ≈  46%

We divide the problem into cases.

Case 1: f(x) > 0.

Then,

\(\begin{array}{rcl} f(f(x)) &=& 4\\ 2f(x) - 15 &=& 4\\ f(x) &=& \dfrac{19}2 \end{array}\)

Case 2: f(x) <= 0.

Then,

\(\begin{array}{rcl} f(f(x)) &=& 4\\ -f(x) + 3&=& 4\\ f(x) &=& -1 \end{array}\)

Then f(x) = 19/2 or f(x) = -1.

We split the problem into 2 cases again.

Case A: x > 0.

Then,

\(2x - 15 = \dfrac{19}2\text{ or }2x - 15 = -1\\ x = \dfrac{49}4\text{ or }x = 7\)

Case B: x <= 0.

Then,

\(-x+3= \dfrac{19}2\text{ or } -x+3 = -1\\ x = -\dfrac{13}2\text{ or }x = 4\text{ (rejected)}\)

Then, solution set = \(\left\{\dfrac{49}4, 7, -\dfrac{13}2\right\}\). So there are a total of 3 solutions.