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Find the total cost of producing 6 widgets.

 Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $23 to produce 2 widgets, $103 to produce 4 widgets, and $631 to produce 10 widgets.

Find the total cost of producing 5 widgets. Widget Wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $15.50 to produce 3 widgets, $23.50 to produce 7 widgets, and $56 to produce 12 widgets.

OK...so we have

a(7)^2 + b(7) + c =  23.50 →   49a + 7b + c = 23.50    (1)

a(3)^2 + b(3) + c =  15.50 →     9a  + 3b + c = 15.50         subtracting the second equation from the first, we have

40a + 4b = 8  → 10a + b = 2    (2)

Also

a(12)^2 + b(12) + c = 56  → 144a + 12b + c = 56   and subtracting (1) from this gives us

95a + 5b = 32.50

And using(2)  we have

95a + 5b = 32.50   (3)

10a + b  =    2.00        multiplying the second equation by -5  and adding  this to (3) ,we have

45a = 22.50      divide both sides by 45    and a = 1/2   and using (2) to find b, we have

10(1/2) + b = 2

5 + b = 2      b = -3

And we can use 9a  + 3b + c = 15.50 to find "c"

9(1/2) + 3(-3) + c = 15.50

9/2 - 9  + c = 15.50

-4.5 + c = 15.50      

c = 20

So our function is

c(x) = (1/2)x^2 - (3)x  + 20

And the cost to produce 5 widgets is = $17.50

Here's the graph of our function

There you go, we are all geniuses!         ♬    

                      ♬                         ♬                                 ♬     

                                  ♬     

                                                                   ♬                                    

Here's my reasoning:

If you are on pedestal 1 the probability of your friend being next to you is 1/23

Same if you are on pedestal 24

If you are on any of the other 22 pedestals the probability is 2/23

Hence the overall weighted probability is (2/24)*(1/23) + (22/24)*(2/23) 

$$\left({\frac{{\mathtt{2}}}{{\mathtt{24}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{23}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{22}}}{{\mathtt{24}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{23}}}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{12}}}} = {\mathtt{0.083\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

Looks like we're calculating the volume of a sphere......you can plug this straight into the on-site calculator.....you should get about  5575.28 cubic units......

I wasn't talking personally Chris.

There are more people in the world than just us you know!    

Well....I wouldn't term either of us as "stupid."  Maybe probability just isn't one of our "stronger" points.....

Doesn't really matter.......if we're correct, someone will let us know (high "probability" of this)...and same "probabilty" if we're not......!!!

Yes i know- My comment was tongue in cheek. 

It is most likely a lot more common for 2 stupid people to think alike.

I wonder what the probablilities are ? 

Yeah....but remember...we thought that about the red/white card problem, too...(as I recall)....!!!

LOL!!!!

Did you register?  What is your username?   

If $1000 is invested at an interest rate of 3.75% per year, compounded quarterly, find the value of the investment after the given number of years. 

i=0.0375/4=0.009375

n years, 4n time periods,

FV=1000(1.009375)⁴ⁿ

FV stands for future value and n stands for number of years.

I don't know Chris, don't they say 2 great minds think alike?  lol

I kinda' suck at probability questions, ND....but I'll take the plunge, here. (I'm sure someone on here will correct me, if I'm wrong!!!........LOL!!)

Let's take a simple example and then apply it to your situation. Instead of 24 people, let's just suppose that we only have four. And let's let each person be represented by a different letter (person 1 =A , person 2 =B, etc.)......Now, we can think of these people as occupying "slots."

Note that there are four letters, A B C D, and there are 24 possible arrangements of these. To see this, note that there are 4 letters possible for slot 1 and three letters possible for slot 2 and two letters possible for slot 3 and one choice for slot 4.  So.....4 x 3 x 2 x 1 = 24.

And let's let you and your friend be "A" and "B"......Note, that there are 2 possible arrangements for both of you.....either AB or BA. And both of you could occupy three different positions...(positions 1,2 .....positions 2,3....or positions 3,4). And note that, the other two people "C" and "D," can be arranged in 2 possible ways for each arrangement.

So, for instance....let's see what it looks like when you and your friend occupy positions 1,2.  We have.....

ABCD or BACD or ABDC or BADC......so, you and your friend are arranged in two possible ways and the other two people are arranged in two possible ways. And since you and your friend could occupy any of three positions, the total arrangements possible where you and your friend are next to each other is given by....

(2 ways for AB) x (2 ways for CD) x (3 possible positions occupied by you and your friend) =  12.

So the probability that you and your friend are next to each other when four people are involved =

(12 arrangements) / (the total arrangements) = 1/2

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

So applying the same reasoning to 24 people...we have 2 arrangements possible for you and your friend in each position x 23 possible positions that you both could occupy x 22! arrangements for the other people in each arrangement. And the total arrangements possible are 24!

So we have  [(2) x 23 x 22!] / 24!  = 1/12

Same as Melody, I believe......however, don't bet your bank account on this!!....we've been known to "b**w a fuse" before .....!!!

Hi Ninja,

Probability is not mystrong suit but I'll give it a shot.

I am assuming that the pedestals are all in a line.  If they are in a circle the answer would be different.

Firstly, how many ways can 24 people be put on 24 pedestals.

24 possiblilities for the 1st one

23 possibilities for the 2nd one 

22 people left for the third one etc

so this is 24! ways.

Now lets think of you and your freind joined together and 2 of the pedestals jointed together.  How many ways can you all be put on the pedestals now?

Well this would be 23! ways except you could be either on the left or on the right of your friend so that would be 23!*2

So I think the probability would be  2*23!/24! = 2/24=1/12

That really does not sound right to me.  Like I said I am not very good at probability.

Your welcome Melody!btw i like sharing things with people out here thats why im going to post these kind of different things alternately!but still ur welcome again!

Thank u everyone for liking my post!

Jboy314 is completely correct.  I just want to stress a point

you use normal trig when you are looking for a length.

You use Inverse trig when you are looking for an angle!

Thank you for sharing Rosala.

The middle 75% means the upper and lower limits are at the points 87.5% and 12.5% respectively (i.e. 25% in total beyond these limits).  

$${\frac{{\mathtt{4}}}{{\mathtt{5}}}}{\mathtt{\,\times\,}}\left({\mathtt{1}}\right){\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{6}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{3}}\right)}{{\mathtt{8}}}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\times\,}}\left({\mathtt{3}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{903}}}{{\mathtt{40}}}} = -{\mathtt{22.575}}$$

so far this is true.

I think this question is in Filipino.

why did he then -903/40 the equal ung nadivide became -22.575

so I think you want to know why this is true.

kaya sa tingin ko nais mong malaman kung bakit ito ay totoo.

arithmetic must be done in a set order of operation. 

It is sometimes referred to as BODMAS or PEDMAS.

[Anyway brackets must be dealt with first], [then exponents (or power Ofs)] then [multiply and divide are equally important] and lastly [addition and subtraction are equally imprortant]

$$\left\{{\frac{{\mathtt{4}}}{{\mathtt{5}}}}{\mathtt{\,\times\,}}\left({\mathtt{1}}\right)\right\}{\mathtt{\,-\,}}\left\{{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{2}}\right)\right\}{\mathtt{\,\small\textbf+\,}}\left\{{\frac{{\mathtt{3}}}{{\mathtt{6}}}}\right\}{\mathtt{\,-\,}}\left\{{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}\left({\mathtt{3}}\right)}{{\mathtt{8}}}}\right\}{\mathtt{\,-\,}}\left\{{\mathtt{6}}{\mathtt{\,\times\,}}\left({\mathtt{3}}\right)\right\}$$

=  $$\left\{{\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right\}{\mathtt{\,-\,}}\left\{{\mathtt{4}}\right\}{\mathtt{\,\small\textbf+\,}}\left\{{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right\}{\mathtt{\,-\,}}\left\{{\frac{{\mathtt{15}}}{{\mathtt{8}}}}\right\}{\mathtt{\,-\,}}\left\{{\mathtt{18}}\right\}$$

=  $$\left\{{\frac{{\mathtt{4}}}{{\mathtt{5}}}}\right\}{\mathtt{\,-\,}}\left\{{\mathtt{4}}\right\}{\mathtt{\,\small\textbf+\,}}\left\{{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right\}{\mathtt{\,-\,}}\left\{{\frac{{\mathtt{15}}}{{\mathtt{8}}}}\right\}{\mathtt{\,-\,}}\left\{{\mathtt{18}}\right\}$$

= $${\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{15}}}{{\mathtt{8}}}}$$

=  $${\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{4}}}{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{15}}}{{\mathtt{8}}}}$$

Now you need to get a common denominator for all those fractions. It will have to be 40

Ngayon ay kailangan mong makakuha ng isang karaniwang denominador para sa lahat ng mga fraction. Ito ay kailangang maging 40

=  $${\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{8}}\right)}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{8}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{20}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{20}}\right)}}{\mathtt{\,-\,}}{\frac{\left({\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}}$$

= $${\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{32}}}{{\mathtt{40}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{20}}}{{\mathtt{40}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{75}}}{{\mathtt{40}}}}$$

= $${\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{32}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}{\mathtt{\,-\,}}{\mathtt{75}}\right)}{{\mathtt{40}}}}$$

= $${\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,-\,}}{\frac{{\mathtt{23}}}{{\mathtt{40}}}}$$        also equals -(22and 23/40)

= $${\mathtt{\,-\,}}{\mathtt{22}}{\mathtt{\,-\,}}{\frac{{\mathtt{23}}}{{\mathtt{40}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{903}}}{{\mathtt{40}}}} = -{\mathtt{22.575}}$$

Does this help explain it?

Tulong na ito ipaliwanag ito ba?

Hi anonymous,

It is very easy to register. If you do it now you can give Alan another point for his solution.  (by giving thumbs up)

                                                            ♬ ♬ MELODY ♬ ♬

There is a calculator on the home site - just add the two together. That is assuming that tanks A and B were full to start with and tank C was empty.

Wow, you are amazing :)

thank you very much, that helped A LOT!

I definitely need to register to this website :)

x2 > 2(x + 4)

You can also do a question like this graphically - which is what I would normally do.

$$x^2>2x+8\\\\
x^2-2x-8>0\\\\
\mbox{You can think of it like this. }\\\\
x^2-2x-8=y \mbox{ Where y is positive}\\\\
y=(x-4)(x+2)\mbox{ Where y is positive}\\\\$$

y=0 when x-4=0 that is when x=4  and also whatn x+2=0  that is when x=-2

So the two roots are x=4 and x=-2.

This is a concave up parabola because the number in front of the x² is +1  (the + indicates concave up)

Do a very quick sketch - you don't need a vertex or any other features.

I'll do a good sketch because with the computer graphing programs it is too difficult to do a rough one.

You can see from the diagram that y is postive when x>4 and when x<-2