Je was op zoek naar: fdist (Engels - Frans)

fdist
loi.f

finv, fdist
inverse.loi.f, loi.f

fdist function
loi.f, fonction

ftest, fdist
ftest, loi.f

fdist(number; degreesfreedom1; degreesfreedom2)
loi.f(nombre;degrés_liberté_1;degrés_liberté_2)

fdist(0.8;8;12) yields 0.61
legacyfdist(0,8;8;12) produit 0,61

=fdist(0.8; 8; 12) yields 0.61.
= loi.f (0,8; 8; 12) donne 0,61.

the fdist() function returns the f-distribution.
la fonction legacyfdist() renvoie la distribution f.

fdist(finv(0.1; 3; 4); 3; 4) equals 0.1
fdist(finv(0.1; 3; 4); 3; 4) vaut 0.1

fdist(number; degrees_freedom 1; degrees_freedom 2)
loi.f (x; degrés_liberté1; degrés_liberté2)

fdist(number; degrees_freedom_1; degrees_freedom_2)
fdist(nombre; degrés_liberté_1; degrés_liberté_2)

\<bookmark_value\>fdist function\</bookmark_value\>
\<bookmark_value\>loi.f, fonction\</bookmark_value\>

the finv() function returns the unique non-negative number x such that fdist(x; r1; r2) = p.
la fonction finv() renvoie l'unique nombre positif x tel que fdist(x; r1; r2) = p.

as the piston rises, when the distributor is in the low position: s 24 +s 20 +s 21 +s 25 =s 18 +s 6 the resultant fb of the forces applied to the distribution box can be written: f b=( s 24 ×hp )+( s 20 ×bp )+( s 21 ×bp )+( s 26 ×hp )−( s 18 ×bp )−( s 6 ×bp )+ f cov +f dist =hp ( s 24 +s 25)+ bp ( s 20 +s 21 −s 18 −s 6)+( hp−bp )( s 7)+( hp−bp )( s 41 −s 40) with fcov=force exerted by the cover on the box fdist=force exerted by the distributor on the box now, from the first equation: s 20 +s 21 −s 18 −s 6 =−s 24 −s 25 hence: f b=( hp−bp )( s 24 +s 25 +s 7 +s 41 −s 40)=( hp−bp ) s t with st being the projected surface area of the view from above of the box.
le couvercle exerce donc un effort sur la boîte de distribution f couv = (hp - bp)(s 7 ) lorsque le distributeur est en position basse, et f couv = (hp - bp)(s 7 ) - (hp - bp)(s 40 ) lorsque le distributeur est en position haute.