Recent content by ulissess

you are right .. for t=4 => f_x=28 and 4 is inside the interval (0,10]; in t=12 => f_x=43 and in t=20 => f_x=28, the same for the max.. you are right ! thank you very much!

i have tried this.. in (0, 10] => f_x=(25+4+9)+10cos(pi/4*t)=38+10cos(pi/4*t) in (10, 18] =>f_x=(25+19+9)+10cos(pi/4*t)=53+10cos(pi/4*t) in (18, 24] =>f_x=(25+4+9)+10cos(pi/4*t)=38+10cos(pi/4*t) for t=0 => f1=48 for t=10 => f2=38 for t=18 => f3=53 for t=24 => f4=48 so for me max=53 and...

yes i know, but i would compare my solution with your solution, i mean i would like to know what max, min absolute you have found.. because i have found max=53 and min=38.. but my book say me different solution (max=54; min=19) so i must understand what i was wrong! (or if the book is wrong)

yes exactly ! , all functions are 0< t<=24

hi, i have these 3 functions [PLAIN]http://img804.imageshack.us/img804/2230/provakd.jpg i must calculate the resultant function, and to find the max, min and the average.. i have tried to calculate it, i have found these values (max=53; min=38; average=43) but the solution in my book are...

i don't understand very well..

y>\frac{z-1}{z+1}x==>z>\frac{y+x}{x-y}

F_Z(z)= P(Z<z) i don't understand find this relation.. maybe for z>1 F_z(Z)= P(Z<z)= P( \frac{x+y}{x-y}<=z>1 ) right? but i don't know how find the right area.. z>1 is all triangle (0,1)*(1,1)*0.5 but i don't know how to continue..

http://img687.imageshack.us/img687/742/unos.jpg 1=F_z(z)+\frac{1}{2}+\frac{z-1}{2(z+1)}=>F_z(z)=\frac{1}{2}-\frac{z-1}{2(z+1)}=\frac{1}{1+z} for z in (-1,1) [PLAIN][PLAIN]http://img217.imageshack.us/img217/1504/trehc.jpg F_z(z)=\frac{1}{2}

1 -1 \frac{z+1}{z-1} http://img683.imageshack.us/img683/7565/duegu.jpg (the black area) 1=F_z(z)+\frac{1}{2}+\frac{z+1}{2(z-1)}=>F_z(z)=\frac{1}{2}-\frac{z+1}{2(z-1)}=\frac{1}{1-z} [PLAIN][PLAIN]http://img694.imageshack.us/img694/5990/unokc.jpg my function is y=\frac{z-1}{z+1}x so...

oh sure .. y=\frac{z-1}{z+1}x with z in (1,+oo) http://img525.imageshack.us/img525/2809/unoj.jpg if (z->+oo) ==> F_z(z)=\int_{0}^{1}xdx=0.5 if z=1 ==>F_z(z)=\int_{0}^{0}1dx=0 F_z(z)=\int_{0}^{1}\frac{z-1}{z+1}dx=\frac{z-1}{2(z+1)} with z in (-oo,-1)...

sorry I'm very shallow... with z in (-oo,-1) http://img256.imageshack.us/img256/1354/unoir.jpg F_z(z)=1*(z+1)/(z-1)*0.5 (area of triangle) with z in (1,+oo) [PLAIN][PLAIN]http://img26.imageshack.us/img26/4328/duep.jpg F_z(z)=1*(z-1)/(z+1)*0.5 (area of triangle)

y=(z+1)/(z-1)*x

http://img689.imageshack.us/img689/6431/55725049.jpg [PLAIN][PLAIN]http://img262.imageshack.us/img262/1655/46800016.jpg i've solved so..

Well, i don't know.. i just try to find a solution. I'm thinking from 3 hours. (z-1)/(z+1) i believe is the curve between 1 and +inf so F(z) use that density but for z<-1 use another pdf between (0,1) so .. I'm lost. Suggestion?