Melody

I never thought that you did post it twice.

I was just inviting you to repost (like you have done here) if you still needed the answer - that is all.  :)

I have now answered you properly on your original post.  

Can I make this a shorter answer?    YES I CAN     

Pat writes all the 7-digit numbers in which all the digits are different and each digit is greater than the one to its right (so the tens digit is greater than the units, the hundreds greater than the tens, and so on). For example, 9,865,320 is one of the numbers that Pat writes down.

(a) How many numbers does Pat write down?

9,8,7,6,5,4,3,2,1,0     3 of the 10 must be removed    10C3 = 120

(b) One of Pat's numbers is chosen at random. What is the probability that the tens digit is a 1?

(9,8,7,6,5,4,3,2) 1,0       3 of  the 8 must be left out     8C3= 56

P(1 in the 10s column) = $${\frac{{\mathtt{56}}}{{\mathtt{120}}}} = {\frac{{\mathtt{7}}}{{\mathtt{15}}}} = {\mathtt{0.466\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

(c) One of Pat's numbers is chosen at random. What is the probability that the middle (thousands) digit is a 5?

(9,8,7,6 one of these four must be left out) 5 ( 4,3,2,1,0 two iof thse 5 must be left out) 4C1*5C2=4*10=40

P(5 is in the middle) =  $${\frac{{\mathtt{40}}}{{\mathtt{120}}}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}} = {\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

Thanks Anon,  

My instructions for reposting were followed to the T so I am going to give some time to your question.

I'll answer you with another question. 

NOTE you are not being asked for the temperature- you are being asked for the overall change in temperature!

If you climb three rungs of a ladder and then go down 3 rungs again where are you

(in relation to where you started?)

Here is a pic that might help you to think about the temperature

Thanks for showing us this Nauseated  

The power of the Web2 calc continues to amaze me  !!      

Hi anon,

You cannot give thumbs up unless you are a member. 

If you were a member you could have give CPhill 5 points:)

Why don't you join up - there are a number of plusses to be had and absolutely no negatives :)

This answer has been modified, 

Thanks CPhill and Alan for letting me know of my careless error :))

(–3/8 Log2(3/8)) + (– 5/8 Log2(5/8)) =

$$\\(-\frac{3}{8}Log_2(\frac{3}{8})) + (-\frac{5}{8} Log_2(\frac{5}{8})) \\\\
=-\frac{1}{8}[3Log_2(\frac{3}{8}) + 5 Log_2(\frac{5}{8})] \\\\
=-\frac{1}{8}[Log_2(\frac{3}{8})^3 + Log_2(\frac{5}{8})^5] \\\\
=-\frac{1}{8}[Log_2((\frac{3}{8})^3*(\frac{5}{8})^5 )] \\\\
=-\frac{1}{8}[Log_2(\frac{(3^3*5^5)}{8^8})] \\\\
=-\frac{1}{8}[Log_2 (3^3*5^5)-Log_2 8^8] \\\\
=-\frac{1}{8}[Log_2 (27*3125)-8Log_2 8] \\\\
=-\frac{1}{8}[Log_2 (84375)-8Log_2 2^3]\\\\
=-\frac{1}{8}[Log_2 (84375)-24Log_2 2]\\\\
=-\frac{1}{8}[Log_2 (84375)-24]\\\\
=3-\frac{[Log_2 (84375)]}{8}\\\\$$

This is the same as Chris's answer below.  :)

The question is finished but if you want an approximation ...

 this can be entered into the web 2 calc like this

3-log(84375,2)/8

$${\mathtt{3}}{\mathtt{\,-\,}}{\frac{{{log}}_{{\mathtt{2}}}{\left({\mathtt{84\,375}}\right)}}{{\mathtt{8}}}} = {\mathtt{0.954\: \!434\: \!002\: \!924\: \!964\: \!7}}$$

Most calcs only do base 10 or base e so this is how you would do it on one of those (uaing change of base rule)

3-(log(84375)/log(2))/8

$${\mathtt{3}}{\mathtt{\,-\,}}{\frac{\left({\frac{{log}_{10}\left({\mathtt{84\,375}}\right)}{{log}_{10}\left({\mathtt{2}}\right)}}\right)}{{\mathtt{8}}}} = {\mathtt{0.954\: \!434\: \!002\: \!924\: \!964\: \!9}}$$

Thanks Alan,

I almost missed this post - how did that happen? 

Thanks for finding this grapher.  It looks really neat.    

I'll add that 3D grapher to our reference material.  I'll add mine too.  I only just thought to do that.   

I didn't think of that Alan, yes there must be a distribution for which B is correct. 

C and D are more than 50% so they will never  be correct!

Thanks Bertie,

This is another that I would like to spend more time on as well. :)