All Answers 358425 Answers

Chris, you made a little mistake when you simplified your fraction. 

The middle would be 7. If you add them all up, you get 15. Then, divide and you get 5. :)

Hi Mellie

Oh yes I saw CPhill answer it too but that was AFTER I made my comment.

I must have got it confused with a similar question. 

Sorry anon :)

I think 1,7,7 unless you want all three different numbers. I just used guess and check.

Yes I get the same as Mellie - I put my answer on the original question :)

I'm sorry: 

The number of possibilities if all the dice look different is 6^4

Only 6 of those will have every cie the sames

So   P(all the same number )=     $${\frac{{\mathtt{6}}}{{{\mathtt{6}}}^{{\mathtt{4}}}}} = {\frac{{\mathtt{1}}}{{\mathtt{216}}}} = {\mathtt{0.004\: \!629\: \!629\: \!629\: \!629\: \!6}}$$

P (they are not all the same)  =   $${\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{6}}}{{{\mathtt{6}}}^{{\mathtt{4}}}}}\right) = {\frac{{\mathtt{215}}}{{\mathtt{216}}}} = {\mathtt{0.995\: \!370\: \!370\: \!370\: \!370\: \!4}}$$

Hey Mellie, what was the address of the post that we all got different answers on?

Pretty sure that wouldn't have an answer. It would be undefined.

It takes a genius.   LOL

Thanks Bertie.   

Thanks a lot!

1/2? Not sure if this was what you were looking for.

I was looking back, and you did repost. We all got different answers. 

It's ok, I trust you. (Not sure about Melody.. Haha) 

Counting the number of outcomes in which four 6-sided dice don't all show the same number would require some pretty delicate casework.

However, counting all the outcomes in which four 6-sided dice do all show the same number is very easy:

there are only 6 ways this can happen, namely all ones, all twos, all threes, all fours, all fives, and all sixes.

So since there are $6^4$ total outcomes, we can conclude that

$$ P(\text{4 dice all show the same number}) = \frac{6}{6^4} = \frac{1}{6^3} = \frac{1}{216}. $$

Therefore, using the principle of complementary probabilities, we can conclude that $$ P(\text{4 dice don't all show the same number}) = 1 - \frac{1}{216} = \boxed{\frac{215}{216}}. $$

Sorry if this didn't come out right, but the answer would be 215/216 :) Hope this helped.

Thanks Alan and Bertie :)

You cannot solve it because there is not = sign.

You can only simplify it.

$$\\\dfrac{(ab^2)^3*b^{-7}}{(a^{-1}*b)^{-2}}\\\\
=\dfrac{a^3b^6b^{-7}}{a^{-1*-2}*b^{-2}}\\\\
=\dfrac{a^3b^6b^{-7}}{a^{2}b^{-2}}\\\\
=\dfrac{a^3b^{6-7}}{a^{2}b^{-2}}\\\\
=\dfrac{a^3b^{-1}}{a^{2}b^{-2}}\\\\
=\dfrac{a^{3-2}b^{-1- -2}}{1}\\\\
=\dfrac{a^{1}b^{1}}{1}\\\\
=ab$$

$$\\V=\frac{(2.21x)}{cos(a)}*\sqrt{-y+tan(a)*x}\\
V^2=\frac{(2.21x)^2}{cos^2(a)}*(-y+tan(a)*x)\\
V^2cos^2a=(2.21x)^2*(-y+tan(a)*x)\\
V^2cos^2a=-(2.21x)^2y+(2.21x)^2tan(a)*x\\
V^2cos^2a+(2.21x)^2y=(2.21x)^2tan(a)*x\\
V^2cos^2a+(2.21x)^2y=(2.21x)^2\sqrt{\frac{1}{cos^2a}-1}*x\\
V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1}{cos^2a}-1)}*x^2\\
V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1-cos^2a}{cos^2a})*x^2\\
V^4cos^6a+2*V^2cos^4a*(2.21x)^2y+(2.21x)^4y^2cos^2a=(2.21x)^4*({1-cos^2a})*x^2\\$$

I am not getting anywhere with this.  Sorry.

Who are we going to believe.  Alan or our friendly neighbourhood troll ?     

Alan has back up evidence... The troll has googly eyes....  Mmm it IS a tough one!!

Anyway I think i will give you thumbs up anyway Mr Troll.  You are entertaining.  :))

Hi   Zacismyname

When you make a post there is a btton in the ribbon called  [Math Formula] 

if you open that and type in   3*sqrt(5) this will be your output.     $${\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}$$

$$\\3^{x+1}=6^{4-x}\\\\
ln[3^{x+1}]=ln[6^{4-x}]\\\\
(x+1)ln3=(4-x)ln6\\\\
xln3+ln3=4ln6-xln6\\\\
xln3+xln6=ln6^4-ln3\\\\
x(ln3+ln6)=ln1296-ln3\\\\
xln18=ln432\\\\
x=\frac{ln432}{ln18}\\\\$$

This answer is identical to Geno's

2(2n+5)=12

4n+10=12

4n=12-10

4n=2

n=2/4

n=1/2

I am almost positive that I saw this exact question already answered today.

If you want to repost PLEASE do it as requested!

3*4 means 3 lots of  4 = 4 +4+4 = 12

3*0.5 means 3 lots of 0.5 = 0.5 + 0.5+0.5 = 1.5 = 1/2 of 3

so 

500*0.5 = 500 lots of 1/2 =  1/2 of 500 = 250

If a camera was reduced by 20%, then the equation would be:

$${\mathtt{X}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{0.2}}\right) = {\mathtt{60}}$$

We simply rearrange to find X.

$${\mathtt{x}} = {\frac{{\mathtt{60}}}{{\mathtt{0.8}}}} \Rightarrow {\mathtt{x}} = {\mathtt{75}}$$

So the original price was £75

Thanks Alan,

This question is really confusing me.    

My answer is also completely different from yours Alan so therefore it is bound to be wrong but maybe you can help clear up my confusion.  I do not understand the wording of the question at all. 

I don't get the 83atm 5000 m altitude - that is jsut for starters.

BUT

The atmospheric pressure decreases with height. Experimentally it is determined that the rate at which the pressure p decreases in function of the height h, is always proportional to p.

Write the differential equation that describes this model and solve general.

$$\\\frac{dP}{dh}=kP\\\\
P=e^{kh}\; +\; c\\\\
check\\
\frac{dP}{dh}=ke^{kh}=kP\qquad $That looks good$\\\\$$

Imagine then p = 1 atm at sea level and p = 0, at 83 atm 5000 m altitude.

When h=0 P=1       1=e^0+c      SO    c=1

$$P=e^{kh}+1$$

p=0  h = total confusion here   p=0, 83atm, 5000m altitude.