Melody

Thanks Chris :)

I used Q to represent older because o is too easily confused with 0.

Q looks a bit like an O that is why I used it.

These types of questions are always much easier if you you letters that are relevant to the question.

I think Heureka's answer is very similarto mine.  He just misinterpreted the last little bit that was being asked for - a very easy error to make :)

Guest's answer is also very similar.  I think the way guest has assigned an easier to work with total is really good.  

It often makes it easier for people to think if there are less letters :)

I have a definite advantage with questions like this - my native language is English.

 It can make a world of difference when interpreting the nuances of language  :)

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

Let 

T = total number of proffessors

M= number of male professors            M=(4/7) T

F = number of female professors         F = (3/7) T

MU (males under 50)

MQ (males over 50)

FU (females under 50)

FQ (females over 50)

(MQ+FQ)=(2/7)T

(MU+FU)=(5/7)T

\(\frac{MQ}{M} =\frac{ 1}{5}\\ \frac{MQ}{ (4/7)T} =\frac{ 1}{5}\\ MQ =\frac{ 1}{5}\times \frac{4T}{7}=\frac{4T}{35}\\ \)

\((MQ+FQ)=\frac{2T}{7}\\ \frac{4T}{35}+FQ=\frac{2T}{7}\\ \frac{4T}{35}+FQ=\frac{10T}{35}\\ FQ=\frac{10T-4T}{35}\\ FQ=\frac{6T}{35}\\ so\\ FU=F-FQ\\ FU=\frac{3T}{7}-\frac{6T}{35}\\ FU=\frac{15T}{35}-\frac{6T}{35}\\ FU=\frac{9T}{35}\\~\\ \frac{FU}{F}=\frac{9T}{35}\div \frac{3T}{7}\\ \frac{FU}{F}=\frac{9T}{35}\times \frac{7}{3T}\\ \frac{FU}{F}=\frac{3}{5}\)

So 3/5 of the female professors are under 50.

Hi SHADES,

I as well as a guest, answered that same question somewhere  else.

Please do not ask the same question in 2 different places!  

It often means that you get 2 identical answers and someone else gets no answer to their question.!

Please follow the guidelines if you want to repost a question!   

Thanks guest #9.    I have edited my answer.     

10^(0.3333(log22+log6.25+log4.2-log7))

\(10^{(0.3333(log22+log6.25+log4.2-log7))}\\~\\ =10^{0.3333log22+0.3333log6.25+0.3333log4.2-0.3333log7)}\\~\\ =10^{0.3333log22}*10^{0.3333log6.25}*10^{0.3333log4.2}*10^{0.3333log7}\\~\\ =10^{log22^{0.3333}}*10^{log6.25^{0.3333}}*10^{log4.2^{0.3333}}*10^{log7^{0.3333}}\\~\\ =22^{0.3333}*6.25^{0.3333}*4.2^{0.3333}*7^{0.3333}\\~\\ =(22*6.25*4.2*7)^{0.3333}\\~\\ =(4042.5)^{0.3333}\\~\\ \approx 15.92562388\)

（x − 1/y）/ （x/y） = x（y-1）/x

\(\frac{（x − 1/y）}{ （x/y）} =\frac{ x（y-1）}{x}\\ Note: x\ne0 \qquad y\ne 0\\ \frac{xy − 1}{y}\div\frac{x}{y} =\frac{ x（y-1）}{x}\\ \frac{xy − 1}{y}\times \frac{y}{x} =\frac{ x（y-1）}{x}\\ \frac{xy − 1}{1}\times \frac{1}{x} =\frac{ 1（y-1）}{1}\\ \frac{xy − 1}{x} =y-1\\ xy − 1=xy-x\\ -1=-x\\ x=1\\ \mbox{So this is the line x=1 except it has a 'hole' at y=0} \)

That was interesting :)

Huh?  I don't do cryptic very well   

That is brilliant MG.  I am super proud of you!!

Don't mind Coldplay.  She is our new  'kid' troll.

She can be really funny when she is at her best!

Hi Shades :)

2 sin2 x - 3 sin x + 1 = 0

\(2 sin^2 x - 3 sin x + 1 = 0\\ let\;\; y=sin^2x\\ 2y^2-3y+1=0\\ 2y^2-2y-y+1=0\\ 2y(y-1)-1(y-1)=0\\ (2y-1)(y-1)=0\\ 2y-1=0 \qquad or \qquad y-1=0\\ 2y=1 \qquad or \qquad y=1\\ y=\frac{1}{2} \qquad or \qquad y=1\\ so\\ sin^2x=\frac{1}{2} \qquad or \qquad sin^2x=1\\ sinx=\pm \frac{1}{\sqrt{2}} \qquad or \qquad sinx=\pm1\\\)

\(x=45^0,135^0,225^0,315^0, 90^0,\;\;or \;\;270^0\)