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[SPAM] Re: [obm-l] Porcentagem
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O pessoal da lista que me corrija se estiver errada.
Aline, tamb=E9m n=E3o consegui resolver a quest=E3o 1, espero a =
solu=E7=E3o dos colegas.
Mas a=ED vai a 2:
Se d =E9 o n=FAmero de litros de leite desnatado e i o n=FAmero de =
litros de leite integral, ent=E3o:
Relativo =E0 gordura -> 0,02d + 0,05i =3D 0,18 (0,18 =E9 3% de 6 =
litros - usando regra de 3)
Relativo ao total de leite -> d + i =3D 6
Resolvendo o sistema temos que:
Da 1=AA equa=E7=E3o -> 5i =3D 18 - 2d
Da 2=AA equa=E7=E3o -> 5i =3D 30 - 5i
E 18 - 2d =3D 30 - 5i -> 3d =3D 12 -> d=3D4
Portanto, letra E, o n=FAmero de litros de leite desnatado =E9 4.
Tamb=E9m era poss=EDvel fazer a equa=E7=E3o relativa a parte do leite =
que n=E3o tem gordura que seria d - 0,02d + i - 0,05i =3D 6, mas d=E1 na =
mesma.
Acho que =E9 isso.
Abra=E7oos.
----- Original Message -----=20
From: Aline=20
To: obm-l=20
Sent: Wednesday, November 07, 2007 5:57 PM
Subject: [obm-l] Porcentagem
Outras duas quest=F5es que estou com d=FAvidas s=E3o:
Se, ao diminuirmos o pre=E7o de certo produto de p% e, em seguida, =
aumentarmos o novo pre=E7o do produto em q%, o pre=E7o volta a ser igual =
ao de antes da diminui=E7=E3o, qual =E9 o valor de 1/p - 1/q?
A) 100
B) 1/100
C) 10
D) 1/10
E) 1
=20
=20
Quantos litros de leite desnatado, com 2% de gordura, devem ser =
misturados com leite integral, com 5% de gordura, para se obter 6 litros =
de leite com 3% de
gordura?
A) 3,2 litros
B) 3,4 litros
C) 3,6 litros
D) 3,8 litros
E) 4 litros
Fico agradecida.
=20
------=_NextPart_000_0024_01C82182.738C9FC0
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML xmlns:o =3D "urn:schemas-microsoft-com:office:office" xmlns:st1 =
=3D=20
"urn:schemas-microsoft-com:office:smarttags"><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
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<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2> O pessoal da lista que me corrija =
se estiver=20
errada.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2> Aline, tamb=E9m n=E3o consegui =
resolver a quest=E3o=20
1, espero a solu=E7=E3o dos colegas.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Mas a=ED vai a 2:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Se d =E9 o n=FAmero de litros de leite =
desnatado e i o=20
n=FAmero de litros de leite integral, ent=E3o:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Relativo =E0 gordura -> 0,02d + =
0,05i =3D=20
0,18 (0,18 =E9 3% de 6 litros - usando =
regra de=20
3)</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Relativo ao total de leite -> d + i =
=3D=20
6</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Resolvendo o sistema temos =
que:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Da 1=AA equa=E7=E3o -> 5i =3D 18 - =
2d</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Da 2=AA equa=E7=E3o -> 5i =3D =
30 - 5i</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>E 18 - 2d =3D 30 - 5i -> =
3d =3D 12 ->=20
d=3D4</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Portanto, letra E, o n=FAmero de litros =
de leite=20
desnatado =E9 4.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Tamb=E9m era poss=EDvel fazer a =
equa=E7=E3o relativa a=20
parte do leite que n=E3o tem gordura que seria d - 0,02d + i - 0,05i =3D =
6, mas d=E1=20
na mesma.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Acho que =E9 isso.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Abra=E7oos.</FONT></DIV>
<BLOCKQUOTE dir=3Dltr=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV=20
style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
<A title=3Doliveira-aline1983@xxxxxxxxxx=20
href=3D"mailto:oliveira-aline1983@xxxxxxxxxx";>Aline</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dobm-l@xxxxxxxxxxxxxx=20
href=3D"mailto:obm-l@xxxxxxxxxxxxxx";>obm-l</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Wednesday, November 07, =
2007 5:57=20
PM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> [obm-l] =
Porcentagem</DIV>
<DIV><BR></DIV>
<DIV style=3D"FONT-SIZE: 12px; FONT-FAMILY: verdana, arial">
<DIV>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial"></SPAN><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial"></SPAN> </P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial"></SPAN> </P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">Outras duas quest=F5es =
que estou com=20
d=FAvidas s=E3o:</SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial"></SPAN> </P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">Se, ao diminuirmos o =
pre=E7o de=20
certo produto de p% e, em seguida, aumentarmos o novo pre=E7o do =
produto em q%,=20
o pre=E7o volta a ser igual ao de antes da diminui=E7=E3o, qual =E9 o =
valor de 1/p =96=20
1/q?<o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">A) =
100<o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">B) =
1/100<o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">C) =
10<o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">D) =
1/10<o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">E) =
1<o:p></o:p></SPAN></P>
<P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: =
justify"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: =
Arial"><o:p> </o:p></SPAN></P>
<P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: =
justify"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: =
Arial"><o:p></o:p></SPAN> </P><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial"><o:p>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">Quantos litros de leite =
desnatado,=20
com 2% de gordura, devem ser misturados com leite integral, com 5% de =
gordura,=20
para se obter <st1:metricconverter ProductID=3D"6 litros" =
w:st=3D"on">6=20
litros</st1:metricconverter> de leite com 3% de<o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: =
Arial">gordura?<o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">A) <st1:metricconverter=20
ProductID=3D"3,2 litros" w:st=3D"on">3,2=20
litros</st1:metricconverter><o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">B) <st1:metricconverter=20
ProductID=3D"3,4 litros" w:st=3D"on">3,4=20
litros</st1:metricconverter><o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">C) <st1:metricconverter=20
ProductID=3D"3,6 litros" w:st=3D"on">3,6=20
litros</st1:metricconverter><o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">D) <st1:metricconverter=20
ProductID=3D"3,8 litros" w:st=3D"on">3,8=20
litros</st1:metricconverter><o:p></o:p></SPAN></P>
<P class=3DMsoNormal=20
style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: justify; =
mso-layout-grid-align: none"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">E) <st1:metricconverter=20
ProductID=3D"4 litros" w:st=3D"on">4=20
litros</st1:metricconverter><o:p></o:p></SPAN></P>
<P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: =
justify"> </P>
<P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: =
justify"> </P>
<P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: =
justify">Fico=20
agradecida.</o:p></SPAN></P>
<P class=3DMsoNormal style=3D"MARGIN: 0cm 0cm 0pt; TEXT-ALIGN: =
justify"><SPAN=20
style=3D"FONT-SIZE: 10pt; FONT-FAMILY: =
Arial"><o:p></o:p></SPAN> </P></DIV></DIV></BLOCKQUOTE></BODY></HTML=
>
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