How to express a group's ratio using individual's ratio? The Next CEO of Stack OverflowFormula...
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How to express a group's ratio using individual's ratio?
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$begingroup$
Can someone help to take a look? Not sure how to start, all numbers are non-negative
express $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ using $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$, would you say $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ is in between $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$?
algebra-precalculus
asked Mar 17 at 16:50
LisaLisa
11815
$endgroup$
add a comment |
$begingroup$
Can someone help to take a look? Not sure how to start, all numbers are non-negative
express $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ using $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$, would you say $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ is in between $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$?
algebra-precalculus
asked Mar 17 at 16:50
LisaLisa
11815
$endgroup$
add a comment |
$begingroup$
Can someone help to take a look? Not sure how to start, all numbers are non-negative
express $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ using $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$, would you say $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ is in between $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$?
algebra-precalculus
asked Mar 17 at 16:50
LisaLisa
11815
$endgroup$
Can someone help to take a look? Not sure how to start, all numbers are non-negative
express $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ using $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$, would you say $frac{S_{1} + S_{2}}{N_{1} + N_{2}}$ is in between $frac{S_{1}}{N_{1}}$ and $frac{S_{2}}{N_{2}}$?
algebra-precalculus
algebra-precalculus
asked Mar 17 at 16:50
LisaLisa
11815
asked Mar 17 at 16:50
LisaLisa
11815
edited Mar 17 at 17:34
Lisa
asked Mar 17 at 16:50
LisaLisa
11815
asked Mar 17 at 16:50
LisaLisa
11815
asked Mar 17 at 16:50
LisaLisa
11815
11815
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint: (I assume all values are non-negative.) The combined expression is a so-called volume-weighted average of the individual expressions.
The weight for $S_i/N_i$ is $N_i$.
A (volume-)weighted average always lies between the smallest and the largest of the terms being averaged. If there are only two such terms, one is the smallest and the other is the largest.
The average can be bounded below by replacing all of the $S_i$ with the smallest one, and bounded above by replacing all of them with the largest one. This produces the desired inequalities.
Note: Here, the volume-weighted average is
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $left(sum_i w_it_iright)/left(sum_i w_iright)$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$.
answered Mar 17 at 17:04
MPWMPW
31k12157
$endgroup$
$begingroup$
how to express the volume-weighted average using the individual expression?
$endgroup$
– Lisa
Mar 17 at 17:34
$begingroup$
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $sum_i w_it_i/sum_i w_i$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$. Will add this to my answer.
$endgroup$
– MPW
Mar 17 at 17:40
add a comment |
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1 Answer
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1 Answer
1
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oldest
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votes
$begingroup$
Hint: (I assume all values are non-negative.) The combined expression is a so-called volume-weighted average of the individual expressions.
The weight for $S_i/N_i$ is $N_i$.
A (volume-)weighted average always lies between the smallest and the largest of the terms being averaged. If there are only two such terms, one is the smallest and the other is the largest.
The average can be bounded below by replacing all of the $S_i$ with the smallest one, and bounded above by replacing all of them with the largest one. This produces the desired inequalities.
Note: Here, the volume-weighted average is
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $left(sum_i w_it_iright)/left(sum_i w_iright)$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$.
answered Mar 17 at 17:04
MPWMPW
31k12157
$endgroup$
$begingroup$
how to express the volume-weighted average using the individual expression?
$endgroup$
– Lisa
Mar 17 at 17:34
$begingroup$
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $sum_i w_it_i/sum_i w_i$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$. Will add this to my answer.
$endgroup$
– MPW
Mar 17 at 17:40
add a comment |
$begingroup$
Hint: (I assume all values are non-negative.) The combined expression is a so-called volume-weighted average of the individual expressions.
The weight for $S_i/N_i$ is $N_i$.
A (volume-)weighted average always lies between the smallest and the largest of the terms being averaged. If there are only two such terms, one is the smallest and the other is the largest.
The average can be bounded below by replacing all of the $S_i$ with the smallest one, and bounded above by replacing all of them with the largest one. This produces the desired inequalities.
Note: Here, the volume-weighted average is
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $left(sum_i w_it_iright)/left(sum_i w_iright)$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$.
answered Mar 17 at 17:04
MPWMPW
31k12157
$endgroup$
$begingroup$
how to express the volume-weighted average using the individual expression?
$endgroup$
– Lisa
Mar 17 at 17:34
$begingroup$
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $sum_i w_it_i/sum_i w_i$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$. Will add this to my answer.
$endgroup$
– MPW
Mar 17 at 17:40
add a comment |
$begingroup$
Hint: (I assume all values are non-negative.) The combined expression is a so-called volume-weighted average of the individual expressions.
The weight for $S_i/N_i$ is $N_i$.
A (volume-)weighted average always lies between the smallest and the largest of the terms being averaged. If there are only two such terms, one is the smallest and the other is the largest.
The average can be bounded below by replacing all of the $S_i$ with the smallest one, and bounded above by replacing all of them with the largest one. This produces the desired inequalities.
Note: Here, the volume-weighted average is
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $left(sum_i w_it_iright)/left(sum_i w_iright)$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$.
answered Mar 17 at 17:04
MPWMPW
31k12157
$endgroup$
Hint: (I assume all values are non-negative.) The combined expression is a so-called volume-weighted average of the individual expressions.
The weight for $S_i/N_i$ is $N_i$.
A (volume-)weighted average always lies between the smallest and the largest of the terms being averaged. If there are only two such terms, one is the smallest and the other is the largest.
The average can be bounded below by replacing all of the $S_i$ with the smallest one, and bounded above by replacing all of them with the largest one. This produces the desired inequalities.
Note: Here, the volume-weighted average is
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $left(sum_i w_it_iright)/left(sum_i w_iright)$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$.
answered Mar 17 at 17:04
MPWMPW
31k12157
edited Mar 17 at 17:41
answered Mar 17 at 17:04
MPWMPW
31k12157
answered Mar 17 at 17:04
MPWMPW
31k12157
answered Mar 17 at 17:04
MPWMPW
31k12157
31k12157
$begingroup$
how to express the volume-weighted average using the individual expression?
$endgroup$
– Lisa
Mar 17 at 17:34
$begingroup$
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $sum_i w_it_i/sum_i w_i$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$. Will add this to my answer.
$endgroup$
– MPW
Mar 17 at 17:40
add a comment |
$begingroup$
how to express the volume-weighted average using the individual expression?
$endgroup$
– Lisa
Mar 17 at 17:34
$begingroup$
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $sum_i w_it_i/sum_i w_i$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$. Will add this to my answer.
$endgroup$
– MPW
Mar 17 at 17:40
$begingroup$
how to express the volume-weighted average using the individual expression?
$endgroup$
– Lisa
Mar 17 at 17:34
$begingroup$
how to express the volume-weighted average using the individual expression?
$endgroup$
– Lisa
Mar 17 at 17:34
$begingroup$
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $sum_i w_it_i/sum_i w_i$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$. Will add this to my answer.
$endgroup$
– MPW
Mar 17 at 17:40
$begingroup$
$[N_1(S_1/N_1) + N_2(S_2/N_2)]/(N_1+N_2)$. In general, a weighted average of terms $t_i$ is $sum_i w_it_i/sum_i w_i$. In your case, the terms $t_i = S_i/N_i$ and the weights $w_i=N_i$. Will add this to my answer.
$endgroup$
– MPW
Mar 17 at 17:40
add a comment |
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