Melody

what is a five digit number contains the digits 1,4,5,6 and 7 but not in that order. arrange these digits so that the resulting number divisible by 4. In how many ways can this be done?

A number is divisable by 4 if the last 2 digits are divisable by 4.

I think the only 2 digit numbers that can be made that are divisable by 4 are 16, 56, 64 and 76  (4 possibilities)

If the last 2 digits are 16 then there are 3! ways ways that the others can be ordered.  That is 6 ways.

The same applies for each of the other final pairs .

So I believe that there are   4*6 = 24 possibilities.

Thanks anon, you are right.  You have answered the question that was asked correctly :)    

However I suspect that the asker had a differen question in mind.  

Anon, If this is not the question that you wanted answered then please try to clarify your question.  

The ONLY answer is c=0

$$\\\frac{1/3}{8}=\frac{x}{2}\\\\
\frac{2/3}{8}=x\\\\
x=\frac{2/3}{8}\\\\$$

$${\frac{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{{\mathtt{8}}}} = {\frac{{\mathtt{1}}}{{\mathtt{12}}}} = {\mathtt{0.083\: \!333\: \!333\: \!333\: \!333\: \!3}}$$    

1/12 of a cup

Your question does not make sense.  Perhaps you mean something like this?

$$\\\displaystyle\lim_{ X\rightarrow 0} \frac{1-cos4X+Xsin3X}{X*X}\\\\$$

If something is raised to a neg power you simply swap it to the other side of the fraction line and change the neg sign to a pos sign.

Just be careful - only the terms with the neg power can be swapped to the other side :))

See if this helps.

$$\\\dfrac{(2-2i)^6(\sqrt{3}+1)^3}{(\sqrt{3}+i)^4}\\\\\\
\dfrac{(2-2i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(\sqrt{3}+i)^4(\sqrt{3}-i)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{\{(\sqrt{3}+i)(\sqrt{3}-i)\}^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(3--1)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(4)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(2)^8}\\\\\\$$

$$\\\dfrac{(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{4}\\\\\\$$

The rest is just painful expansion.  Maybe you do not need to go any further.

(1-i)^6 will work out the easiest.  Use binomial expansions. 

If you need help expanding I suggest you ask for it.   :) 

I am glad that you are back TR  

your class average on a test was a 70 with a standard deviation of 3 points.What percent of the class fell between 67 and 76 points 

I am going to split this into 2 parts.  Below the mean and above the mean.

70-3=67           so 67 is one standard deviation below the mean

between -1 and +1 SD  from the mean lie 64% of all scores.  Half of this is 34%

70+2*3=76      and 76 is 2 standard deviations above the mean.

between -2 and +2 SD  from the mean lie 95% of all scores.  Half of this is 47.5%

Total prob of lying in desired region = 34+47.5 = 81.5%    approximately.

I do not understand your question :/