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This is what I have come up with.  :/   [no guarantees that it is correct though]

Where on Earth are you digging these questions up from Mellie. :))

Now I have to work out how to calculate this without doing every bit separately.

Got any ideas Alan?  Do you know how to get the web2 calc to do the numerator?

$$\displaystyle\sum_{x=1}^{40}\left(\frac{39!}{(40-x)!}\times(52-x)!\right)\div 52!$$

-2(2x+3)+6x-9.                                                                                                                                                           (1      -4x+-6+6x-9

       2x+-15

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1/2+2/5t-1=1/5t+t                                                                                                                                                    (2

  -1/2+2/5t=1.2t

          -2\5t= -2/5t

    -2/1* -1/2=.8t*-2/1

               t=-1.6

What do you mean by the end? We're you supposed to give us the money you had prior to buying the apples? If not, I will just calculate the amount you spend on the five apples. It's just multiplication. 

$${\mathtt{6.5}}{\mathtt{\,\times\,}}{\mathtt{5}} = {\frac{{\mathtt{65}}}{{\mathtt{2}}}} = {\mathtt{32.5}}$$

£32.50

Basically, the measures are like different variables because they have different values. So, first we want to add the meters.

2m+5m-3m=4m

Then, do the centimeters.

40cm+25cm-80cm=-15cm. 

You will get 3.85 m in total

Expected value = (sum of weight*number)/(sum of weights)

So, for part 1:  Expected value = (6*1 + 2*3)/(6 + 2)  =  12/8  =  3/2

Part 2:  Expected value =  (6*1 + 3*3)/(6 + 3) = 15/9 = 5/3

 See if you can now do the other parts.

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$${\mathtt{207\,000\,600}}{\mathtt{\,\small\textbf+\,}}{\mathtt{207\,000\,600}} = {\mathtt{414\,001\,200}}$$

$${{\mathtt{13}}}^{{\mathtt{123}}} = {\mathtt{103\,521\,923\,711\,427\,330\,139\,508\,660\,561\,111\,399\,310\,356\,231\,562\,909\,825\,053\,544\,787\,423\,528\,887\,795\,603\,512\,927\,942\,762\,680\,653\,876\,132\,399\,975\,884\,814\,583\,882\,073\,552\,996\,857\,463\,797}}$$

x = pi is also a solution

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Who would like to add their birthday to the list ?   

how do I solve for x in, sin2x=2sinx when in the intervals of [0,2pi)?

$$\\2sinxcosx=2sinx\\
cosx=1\\
x=0$$

what is the question?

What fun happened can I hear the long story?

Thanks Maths.......God? 

http://www.binaryhexconverter.com/hex-to-decimal-converter

@@ End of Day Wrap    Sat 18/4/15 Sydney,   Australia Time   9:40pm   ♪ ♫

Good evening   

Great answers today from Alan, Trincent, Zacismyname, Badinage, CPhill, Brittany and Zegroes. Thank you :)

Rosala is back.  She has been away at a wedding the second one in about a month.  I think she only came home to change her clothes      Anyway, she is back home and back with us now.  Welcome back Rosala.  

Interest posts:

FTJ means "For the Juniors"

Sun 19/4/15

FTJ means "For the Juniors"

Thanks Badinage :)

I am going to try and do it by hand

$$\\(x-2)^{x-3}<(x-2)^{5-x}\\\\
(x-3)log(x-2)<(5-x)log(x-2)\\\\
(x-3)log(x-2)-(5-x)log(x-2)<0\\\\
(x-3-5+x)log(x-2)<0\\\\
(2x-8)log(x-2)<0\\\\
(x-4)log(x-2)<0$$

this will be true if one of the factors is negative and the other is positive.

log(x-2)>0 when x-2>1    that is when x>3

x-4>0                                    when   x>4  

so

If x4 then both are negative.

BUT   the goldilocks moment comes

when  x is between 3 and 4 because then x-4 is neg and log(x-2) is positive

so

This will be true when     3<x<4

Thanks Alan,

So you want the blue graph to be UNDER the green graph (because it is less than)

this appears to happen between x=3 and x=4  :))

3<x<4

I have added this info to our reference material sticky thread  (In the calculator section)    

Melody, the original thread was back here:

☛ ☚

That is really confusing Alan but I think I finally understand what you have done.

It would be very slow to use this on a regular basis but maybe practice would increase speed:/

It would be SOOO much easier to understand and do if the constant coefficients were just a,b,c etc   

It amazes me sometimes that people can be so very clever to program these things and yet they make the most obvious errors or complications.   :/

Thanks very much for taking the time to look at this for me :))

Plotting a graph often helps:

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3 is a solution

$${{\mathtt{3}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{10}}$$

(It's the only real solution)

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Bertie, (or anyone with statistics knowledge) can you help please ?

Pretty please with sprinkles on top :)))

I do not know if this is correct.

I don't know what to do with the 0.04 either