Mixing
A system became $20%$ slower than earlier time $T$ should be modeled as $T*1.2$ or $frac{T}{0.8}$
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I am bit confused in the following question as already states in the topic
A system became $20%$ slower than earlier time $T$ it used to take to complete an operation. Should it be modeled as $T*1.2$ or $frac{T}{0.8}$?
Actually what each of these representation imply? i.e, $T*1.2$ and $frac{T}{0.8}$? And which would be correct modeling?
Also, What if the system became $20%$ faster? I have been correctly solving modeling $frac{T}{1.2}$ but I am curious what $T*0.8$ would mean in this case?
discrete-mathematics
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
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add a comment |
$begingroup$
I am bit confused in the following question as already states in the topic
A system became $20%$ slower than earlier time $T$ it used to take to complete an operation. Should it be modeled as $T*1.2$ or $frac{T}{0.8}$?
Actually what each of these representation imply? i.e, $T*1.2$ and $frac{T}{0.8}$? And which would be correct modeling?
Also, What if the system became $20%$ faster? I have been correctly solving modeling $frac{T}{1.2}$ but I am curious what $T*0.8$ would mean in this case?
discrete-mathematics
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
$endgroup$
2
$begingroup$
So let's say that a system has as speed, defined as $$ S = frac{1}{T} $$ If the system becomes $20~%$ slower, $S$ is reduced by a factor of $0.8$: $$ 0.8 S = frac{0.8}{T} $$ I think from this you can already see the answer.
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– Matti P.
Feb 1 at 9:26
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Thanks @MattiP. What $T*1.2$ imply?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:32
$begingroup$
For that, you can use what you learned from my comment and apply it.
$endgroup$
– Matti P.
Feb 1 at 9:44
$begingroup$
@MattiP. $16.7%$ decrease in speed?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:46
add a comment |
$begingroup$
I am bit confused in the following question as already states in the topic
A system became $20%$ slower than earlier time $T$ it used to take to complete an operation. Should it be modeled as $T*1.2$ or $frac{T}{0.8}$?
Actually what each of these representation imply? i.e, $T*1.2$ and $frac{T}{0.8}$? And which would be correct modeling?
Also, What if the system became $20%$ faster? I have been correctly solving modeling $frac{T}{1.2}$ but I am curious what $T*0.8$ would mean in this case?
discrete-mathematics
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
$endgroup$
I am bit confused in the following question as already states in the topic
A system became $20%$ slower than earlier time $T$ it used to take to complete an operation. Should it be modeled as $T*1.2$ or $frac{T}{0.8}$?
Actually what each of these representation imply? i.e, $T*1.2$ and $frac{T}{0.8}$? And which would be correct modeling?
Also, What if the system became $20%$ faster? I have been correctly solving modeling $frac{T}{1.2}$ but I am curious what $T*0.8$ would mean in this case?
discrete-mathematics
discrete-mathematics
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
asked Feb 1 at 9:24
Mr.Sigma.Mr.Sigma.
18311
18311
2
$begingroup$
So let's say that a system has as speed, defined as $$ S = frac{1}{T} $$ If the system becomes $20~%$ slower, $S$ is reduced by a factor of $0.8$: $$ 0.8 S = frac{0.8}{T} $$ I think from this you can already see the answer.
$endgroup$
– Matti P.
Feb 1 at 9:26
$begingroup$
Thanks @MattiP. What $T*1.2$ imply?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:32
$begingroup$
For that, you can use what you learned from my comment and apply it.
$endgroup$
– Matti P.
Feb 1 at 9:44
$begingroup$
@MattiP. $16.7%$ decrease in speed?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:46
add a comment |
2
$begingroup$
So let's say that a system has as speed, defined as $$ S = frac{1}{T} $$ If the system becomes $20~%$ slower, $S$ is reduced by a factor of $0.8$: $$ 0.8 S = frac{0.8}{T} $$ I think from this you can already see the answer.
$endgroup$
– Matti P.
Feb 1 at 9:26
$begingroup$
Thanks @MattiP. What $T*1.2$ imply?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:32
$begingroup$
For that, you can use what you learned from my comment and apply it.
$endgroup$
– Matti P.
Feb 1 at 9:44
$begingroup$
@MattiP. $16.7%$ decrease in speed?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:46
2
2
$begingroup$
So let's say that a system has as speed, defined as $$ S = frac{1}{T} $$ If the system becomes $20~%$ slower, $S$ is reduced by a factor of $0.8$: $$ 0.8 S = frac{0.8}{T} $$ I think from this you can already see the answer.
$endgroup$
– Matti P.
Feb 1 at 9:26
$begingroup$
So let's say that a system has as speed, defined as $$ S = frac{1}{T} $$ If the system becomes $20~%$ slower, $S$ is reduced by a factor of $0.8$: $$ 0.8 S = frac{0.8}{T} $$ I think from this you can already see the answer.
$endgroup$
– Matti P.
Feb 1 at 9:26
$begingroup$
Thanks @MattiP. What $T*1.2$ imply?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:32
$begingroup$
Thanks @MattiP. What $T*1.2$ imply?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:32
$begingroup$
For that, you can use what you learned from my comment and apply it.
$endgroup$
– Matti P.
Feb 1 at 9:44
$begingroup$
For that, you can use what you learned from my comment and apply it.
$endgroup$
– Matti P.
Feb 1 at 9:44
$begingroup$
@MattiP. $16.7%$ decrease in speed?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:46
$begingroup$
@MattiP. $16.7%$ decrease in speed?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:46
add a comment |
1 Answer
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Imagine that the task involved is to run (or walk) a distance of one mile.
If you run at speed $S$ miles per hour then the time taken to complete the mile is $T=frac 1 S$ hours. If you now run $20%$ slower then your speed is now $S'=0.8S$ and your time to complete the mile is now
$T'=frac{1}{S'} = frac{1}{0.8S} = frac{T}{0.8}$
If instead you have $T'=1.2T$ then we would say you took $20%$ longer - which is different from running $20%$ slower.
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
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add a comment |
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1 Answer
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$begingroup$
Imagine that the task involved is to run (or walk) a distance of one mile.
If you run at speed $S$ miles per hour then the time taken to complete the mile is $T=frac 1 S$ hours. If you now run $20%$ slower then your speed is now $S'=0.8S$ and your time to complete the mile is now
$T'=frac{1}{S'} = frac{1}{0.8S} = frac{T}{0.8}$
If instead you have $T'=1.2T$ then we would say you took $20%$ longer - which is different from running $20%$ slower.
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
$endgroup$
add a comment |
$begingroup$
Imagine that the task involved is to run (or walk) a distance of one mile.
If you run at speed $S$ miles per hour then the time taken to complete the mile is $T=frac 1 S$ hours. If you now run $20%$ slower then your speed is now $S'=0.8S$ and your time to complete the mile is now
$T'=frac{1}{S'} = frac{1}{0.8S} = frac{T}{0.8}$
If instead you have $T'=1.2T$ then we would say you took $20%$ longer - which is different from running $20%$ slower.
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
$endgroup$
add a comment |
$begingroup$
Imagine that the task involved is to run (or walk) a distance of one mile.
If you run at speed $S$ miles per hour then the time taken to complete the mile is $T=frac 1 S$ hours. If you now run $20%$ slower then your speed is now $S'=0.8S$ and your time to complete the mile is now
$T'=frac{1}{S'} = frac{1}{0.8S} = frac{T}{0.8}$
If instead you have $T'=1.2T$ then we would say you took $20%$ longer - which is different from running $20%$ slower.
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
$endgroup$
Imagine that the task involved is to run (or walk) a distance of one mile.
If you run at speed $S$ miles per hour then the time taken to complete the mile is $T=frac 1 S$ hours. If you now run $20%$ slower then your speed is now $S'=0.8S$ and your time to complete the mile is now
$T'=frac{1}{S'} = frac{1}{0.8S} = frac{T}{0.8}$
If instead you have $T'=1.2T$ then we would say you took $20%$ longer - which is different from running $20%$ slower.
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
answered Feb 1 at 9:57
gandalf61gandalf61
9,232825
9,232825
add a comment |
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$begingroup$
So let's say that a system has as speed, defined as $$ S = frac{1}{T} $$ If the system becomes $20~%$ slower, $S$ is reduced by a factor of $0.8$: $$ 0.8 S = frac{0.8}{T} $$ I think from this you can already see the answer.
$endgroup$
– Matti P.
Feb 1 at 9:26
$begingroup$
Thanks @MattiP. What $T*1.2$ imply?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:32
$begingroup$
For that, you can use what you learned from my comment and apply it.
$endgroup$
– Matti P.
Feb 1 at 9:44
$begingroup$
@MattiP. $16.7%$ decrease in speed?
$endgroup$
– Mr.Sigma.
Feb 1 at 9:46