Melody

\(S_n=\frac{n}{2}(a+L) = \frac{20002}{2}(1+20002)=10001*20003 = 200\;050\;003\)

Please resubmit your question and input your LaTex  properly, via the LaTex box (which can be found in the ribbon.)

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If you are too lazy to do this then why should anyone be bothered to answer you.

I have closed this question.

There are 100 numbers, that is the n(S)   the number in the sample space.

Prime actors of 18 are 2,3 .  The product of prime factors is 6

there are  50 factors of 2, all of those are reducible

There are  33 factors of 3 , all are reducible

There are  16 factors of 6, these have been counted twice

so number of desirable events is    50+33-16 = 67  

Prob = 67/100

\(p(x)=(x-1)(x-3)Q(x)+ax+b\\ \text{So the remainder when p(x) is divided by }(x-1)(x-3) \;\;is\;\; ax+b\\~\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax+b}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax-a+b+a}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+a+\frac{b+a}{x-1}\\ \qquad \text {But we already know that the remainder is 3 so}\\ \qquad a+b=3\qquad (1)\\~\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{ax+b}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+a+\frac{b+3a}{x-3}\\ \qquad \text {But we already know that the remainder is 8 so}\\ \qquad b+3a=8\qquad \\ \qquad b+a+2a=8\qquad \\ \qquad 3+2a=8\qquad \\ \qquad a=2.5 \qquad \\~\\ \qquad b=3-a\\ \qquad b=3-2.5\\\qquad b=0.5 \qquad \\~\\ \)

So when  p(x) is divided by (x-1)(x-3)  the remainder is   2.5x+0.5

I have not checked that this is correct,   lets see

\(p(x)=(x-1)(x-3)Q(x)+2.5x+0.5\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x+0.5}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x-2.5+3}{x-1}\\ \frac{p(x)}{(x-1)}=(x-3)Q(x)+2.5+\frac{3}{x-1}\qquad Cool!\\~\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5x+0.5}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5x-7.5+0.5+7.5}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5(x-3)+0.5+7.5}{x-3}\\ \frac{p(x)}{(x-3)}=(x-1)Q(x)+2.5+\frac{8}{x-3}\qquad Double\;\;Cool!\\ \)

LaTex:

p(x)=(x-1)(x-3)Q(x)+ax+b\\
\text{So the remainder when p(x) is divided by }(x-1)(x-3) \;\;is\;\; ax+b\\~\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax+b}{x-1}\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{ax-a+b+a}{x-1}\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+a+\frac{b+a}{x-1}\\
\qquad \text {But we already know that the remainder is 3 so}\\
\qquad  a+b=3\qquad (1)\\~\\

\frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{ax+b}{x-3}\\
\frac{p(x)}{(x-3)}=(x-1)Q(x)+a+\frac{b+3a}{x-3}\\
\qquad \text {But we already know that the remainder is 8 so}\\
\qquad  b+3a=8\qquad \\
\qquad  b+a+2a=8\qquad \\
\qquad  3+2a=8\qquad \\
\qquad  a=2.5 \qquad \\~\\
\qquad  b=3-a\\ \qquad b=3-2.5\\\qquad b=0.5 \qquad \\~\\

p(x)=(x-1)(x-3)Q(x)+2.5x+0.5\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x+0.5}{x-1}\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+\frac{2.5x-2.5+3}{x-1}\\
\frac{p(x)}{(x-1)}=(x-3)Q(x)+2.5+\frac{3}{x-1}\qquad Cool!\\~\\

\frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5x+0.5}{x-3}\\
\frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5x-7.5+0.5+7.5}{x-3}\\
\frac{p(x)}{(x-3)}=(x-1)Q(x)+\frac{2.5(x-3)+0.5+7.5}{x-3}\\
\frac{p(x)}{(x-3)}=(x-1)Q(x)+2.5+\frac{8}{x-3}\qquad Double\;\;Cool!\\
 

Your order matters is correct but if you were trying to help guest, well you didn't, because you did not explain where your numbers came from/

If you were just doing it for yourself.... well... ok

Your order not matter is NOT correct because when you worked out the totoal number of possible rolls you assumed order DID matter.

Hi Straight and guest.

It makes it ieasier if you think of each die being a different colour.

1*2*6=12         3!= 6 ways

1*3*4=12               6 ways

2*2*3=12              3 ways

that is it I think

6+6+3 = 15 ways

6*6*6 = 216

so the prob will be    15 / 215  =     5 / 72

I have not done this question but this is how I would try tackling it.

work out formula for the distance between (0,0)  and  (  x, (x^2-18)/sqrt2   )

Then use calculus to minimize it.

Before hertrip to Hong Kong, Oga exchanged C$500 for HK$3287. About how many Hong Kongdollars did Oga receive for each Canadian dollar?

You want to know     Hong Kong dollars  PER  Canadian dollar

So you take the Hong Kong amount and DIVIDE it by the canadian amount.

Please put your latex into a latex box (on the ribon) so that it displays properly.

I have not thought this question out well.   I am not cllaiming it is exactly what you are asked for.

I am not even sure it is correct, Iam too tired to think it through well. 

But perhaps what I have done will give you some ideas.

A trading day is usually 6 hours long.  So the horizonal axis goes from  0 to  6/t    Domain   0 <= t <= 6/k 

If k >1.5 then the lines won't cross before the end of the day and    Range will be       0 < d < 3(6/k)+4  that is range :  0

If k <=1.5 then the lines will cross before the end of the day and    Range will be       0 < d < 4 (6/k)  that is range :  0