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what does it mean to reject a hypothesis?
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This is a simple problem I cannot understand. The task is to solve:
$$frac23 times |5-2x| - frac12 = 5$$
- As a first step, I isolate the absolute value, like this:
beginalign
2/3 times |5-2x| &= 5 + 1/2\
2/3 times |5-2x| &= 11/2\
|5-2x| &= 11/2 ÷ 2/3 \
|5-2x| &= 33/4\
|5-2x| &= 8.25
endalign
2. Now, $|5-2x|$ can be two things:
$$
|5-2x| =begincases 5-2x & x geq 2.5\
5+2x & x < 2.5.
endcases
$$
- Let's solve the equation $|5-2x| = 8.25$
if $x geq 2.5$, then:
beginalign
5-2x &= 8.25\
-2x &= 3.25\
x &= -1.625
endalign
But we said that $x geq 2.5$!
I get a similar contradiction if I calculte the other conditional: $x < 2.5$
beginalign
-5+2x &= 8.25\
2x &= 13.25\
x &= 6.625
endalign
which is not smaller than $2.5$.
As far as I know, in such situations, we tend "reject the solution." But what does it mean for a solution not to confirm to a hypothesis?
(The problem is from Stitz & Zeager (2013) Precalculus, exercise: 2.2.1/8.
I apologise for not being able to use nice formatting.)
self-learning absolute-value
edited Jul 26 at 15:47
Foobaz John
18k41245
asked Jul 26 at 15:39
malasi
204
add a comment |Â
up vote
0
down vote
favorite
This is a simple problem I cannot understand. The task is to solve:
$$frac23 times |5-2x| - frac12 = 5$$
- As a first step, I isolate the absolute value, like this:
beginalign
2/3 times |5-2x| &= 5 + 1/2\
2/3 times |5-2x| &= 11/2\
|5-2x| &= 11/2 ÷ 2/3 \
|5-2x| &= 33/4\
|5-2x| &= 8.25
endalign
2. Now, $|5-2x|$ can be two things:
$$
|5-2x| =begincases 5-2x & x geq 2.5\
5+2x & x < 2.5.
endcases
$$
- Let's solve the equation $|5-2x| = 8.25$
if $x geq 2.5$, then:
beginalign
5-2x &= 8.25\
-2x &= 3.25\
x &= -1.625
endalign
But we said that $x geq 2.5$!
I get a similar contradiction if I calculte the other conditional: $x < 2.5$
beginalign
-5+2x &= 8.25\
2x &= 13.25\
x &= 6.625
endalign
which is not smaller than $2.5$.
As far as I know, in such situations, we tend "reject the solution." But what does it mean for a solution not to confirm to a hypothesis?
(The problem is from Stitz & Zeager (2013) Precalculus, exercise: 2.2.1/8.
I apologise for not being able to use nice formatting.)
self-learning absolute-value
edited Jul 26 at 15:47
Foobaz John
18k41245
asked Jul 26 at 15:39
malasi
204
1
Why do you say $|5-2x|=5-2x$ when $xge 2.5$?
– Lord Shark the Unknown
Jul 26 at 15:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is a simple problem I cannot understand. The task is to solve:
$$frac23 times |5-2x| - frac12 = 5$$
- As a first step, I isolate the absolute value, like this:
beginalign
2/3 times |5-2x| &= 5 + 1/2\
2/3 times |5-2x| &= 11/2\
|5-2x| &= 11/2 ÷ 2/3 \
|5-2x| &= 33/4\
|5-2x| &= 8.25
endalign
2. Now, $|5-2x|$ can be two things:
$$
|5-2x| =begincases 5-2x & x geq 2.5\
5+2x & x < 2.5.
endcases
$$
- Let's solve the equation $|5-2x| = 8.25$
if $x geq 2.5$, then:
beginalign
5-2x &= 8.25\
-2x &= 3.25\
x &= -1.625
endalign
But we said that $x geq 2.5$!
I get a similar contradiction if I calculte the other conditional: $x < 2.5$
beginalign
-5+2x &= 8.25\
2x &= 13.25\
x &= 6.625
endalign
which is not smaller than $2.5$.
As far as I know, in such situations, we tend "reject the solution." But what does it mean for a solution not to confirm to a hypothesis?
(The problem is from Stitz & Zeager (2013) Precalculus, exercise: 2.2.1/8.
I apologise for not being able to use nice formatting.)
self-learning absolute-value
edited Jul 26 at 15:47
Foobaz John
18k41245
asked Jul 26 at 15:39
malasi
204
This is a simple problem I cannot understand. The task is to solve:
$$frac23 times |5-2x| - frac12 = 5$$
- As a first step, I isolate the absolute value, like this:
beginalign
2/3 times |5-2x| &= 5 + 1/2\
2/3 times |5-2x| &= 11/2\
|5-2x| &= 11/2 ÷ 2/3 \
|5-2x| &= 33/4\
|5-2x| &= 8.25
endalign
2. Now, $|5-2x|$ can be two things:
$$
|5-2x| =begincases 5-2x & x geq 2.5\
5+2x & x < 2.5.
endcases
$$
- Let's solve the equation $|5-2x| = 8.25$
if $x geq 2.5$, then:
beginalign
5-2x &= 8.25\
-2x &= 3.25\
x &= -1.625
endalign
But we said that $x geq 2.5$!
I get a similar contradiction if I calculte the other conditional: $x < 2.5$
beginalign
-5+2x &= 8.25\
2x &= 13.25\
x &= 6.625
endalign
which is not smaller than $2.5$.
As far as I know, in such situations, we tend "reject the solution." But what does it mean for a solution not to confirm to a hypothesis?
(The problem is from Stitz & Zeager (2013) Precalculus, exercise: 2.2.1/8.
I apologise for not being able to use nice formatting.)
self-learning absolute-value
edited Jul 26 at 15:47
Foobaz John
18k41245
asked Jul 26 at 15:39
malasi
204
edited Jul 26 at 15:47
Foobaz John
18k41245
edited Jul 26 at 15:47
Foobaz John
18k41245
edited Jul 26 at 15:47
Foobaz John
18k41245
18k41245
asked Jul 26 at 15:39
malasi
204
asked Jul 26 at 15:39
malasi
204
asked Jul 26 at 15:39
malasi
204
204
1
Why do you say $|5-2x|=5-2x$ when $xge 2.5$?
– Lord Shark the Unknown
Jul 26 at 15:43
add a comment |Â
1
Why do you say $|5-2x|=5-2x$ when $xge 2.5$?
– Lord Shark the Unknown
Jul 26 at 15:43
1
1
Why do you say $|5-2x|=5-2x$ when $xge 2.5$?
– Lord Shark the Unknown
Jul 26 at 15:43
Why do you say $|5-2x|=5-2x$ when $xge 2.5$?
– Lord Shark the Unknown
Jul 26 at 15:43
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$$5-2xge 0iff |5-2x|=5-2x$$ but $$5-2xge0iff 2xle 5,$$
unlike what you wrote.
In this particular exercise, no hypothesis is rejected, so it doesn't illustrate the concept.
answered Jul 26 at 15:46
Yves Daoust
110k665203
I didn't know that multiplying an inequality by -1 changes the direction of the inequality sign. Thank you so much for pointing that out, Yves! Also, please, let me know if I should edit or do something with the post, because (I realise) it's really not a question about hypothesis rejection. Thank you.
– malasi
Jul 26 at 17:13
@malasi: I did'nt work this out with a multiplication but by changing members.
– Yves Daoust
Jul 26 at 17:44
what do you mean by "changing members?" I'm confused now.
– malasi
Jul 26 at 17:52
I got it, you simply added +2x to the inequality 5-2x ≥ 0. I had somehow gotten confused by the order of members. Thanks for the notice.
– malasi
Jul 26 at 20:13
add a comment |Â
up vote
1
down vote
In addition to what has been said, note that we can solve the equation by noting that
$$
|x|=aiff x=pm a
$$
In particular
$$
|5-2x|=8.25iff 5-2x=8.25 quad textor quad 5-2x=-8.25
$$
which is perhaps easier to solve.
answered Jul 26 at 15:50
Foobaz John
18k41245
add a comment |Â
up vote
0
down vote
A good way to solve this kind of equation is consider two cases
1) For $5-2xge 0 iff xle frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (5-2x) - 1/2 = 5 iff 20-8x-3=30 \iff 8x=-13 iff x=-frac138$$
that solution is acceptable since it is consistent to the assumption $xle frac25$.
2) For $5-2x< 0 iff x> frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (2x-5) - 1/2 = 5 iff 8x-20-3=30 \iff 8x=53 iff x=frac538$$
which is also acceptable.
As an example with some solution to reject refer to Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$.
answered Jul 26 at 15:45
gimusi
65k73583
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$5-2xge 0iff |5-2x|=5-2x$$ but $$5-2xge0iff 2xle 5,$$
unlike what you wrote.
In this particular exercise, no hypothesis is rejected, so it doesn't illustrate the concept.
answered Jul 26 at 15:46
Yves Daoust
110k665203
I didn't know that multiplying an inequality by -1 changes the direction of the inequality sign. Thank you so much for pointing that out, Yves! Also, please, let me know if I should edit or do something with the post, because (I realise) it's really not a question about hypothesis rejection. Thank you.
– malasi
Jul 26 at 17:13
@malasi: I did'nt work this out with a multiplication but by changing members.
– Yves Daoust
Jul 26 at 17:44
what do you mean by "changing members?" I'm confused now.
– malasi
Jul 26 at 17:52
I got it, you simply added +2x to the inequality 5-2x ≥ 0. I had somehow gotten confused by the order of members. Thanks for the notice.
– malasi
Jul 26 at 20:13
add a comment |Â
up vote
1
down vote
accepted
$$5-2xge 0iff |5-2x|=5-2x$$ but $$5-2xge0iff 2xle 5,$$
unlike what you wrote.
In this particular exercise, no hypothesis is rejected, so it doesn't illustrate the concept.
answered Jul 26 at 15:46
Yves Daoust
110k665203
I didn't know that multiplying an inequality by -1 changes the direction of the inequality sign. Thank you so much for pointing that out, Yves! Also, please, let me know if I should edit or do something with the post, because (I realise) it's really not a question about hypothesis rejection. Thank you.
– malasi
Jul 26 at 17:13
@malasi: I did'nt work this out with a multiplication but by changing members.
– Yves Daoust
Jul 26 at 17:44
what do you mean by "changing members?" I'm confused now.
– malasi
Jul 26 at 17:52
I got it, you simply added +2x to the inequality 5-2x ≥ 0. I had somehow gotten confused by the order of members. Thanks for the notice.
– malasi
Jul 26 at 20:13
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$5-2xge 0iff |5-2x|=5-2x$$ but $$5-2xge0iff 2xle 5,$$
unlike what you wrote.
In this particular exercise, no hypothesis is rejected, so it doesn't illustrate the concept.
answered Jul 26 at 15:46
Yves Daoust
110k665203
$$5-2xge 0iff |5-2x|=5-2x$$ but $$5-2xge0iff 2xle 5,$$
unlike what you wrote.
In this particular exercise, no hypothesis is rejected, so it doesn't illustrate the concept.
answered Jul 26 at 15:46
Yves Daoust
110k665203
edited Jul 26 at 15:58
answered Jul 26 at 15:46
Yves Daoust
110k665203
answered Jul 26 at 15:46
Yves Daoust
110k665203
answered Jul 26 at 15:46
Yves Daoust
110k665203
110k665203
I didn't know that multiplying an inequality by -1 changes the direction of the inequality sign. Thank you so much for pointing that out, Yves! Also, please, let me know if I should edit or do something with the post, because (I realise) it's really not a question about hypothesis rejection. Thank you.
– malasi
Jul 26 at 17:13
@malasi: I did'nt work this out with a multiplication but by changing members.
– Yves Daoust
Jul 26 at 17:44
what do you mean by "changing members?" I'm confused now.
– malasi
Jul 26 at 17:52
I got it, you simply added +2x to the inequality 5-2x ≥ 0. I had somehow gotten confused by the order of members. Thanks for the notice.
– malasi
Jul 26 at 20:13
add a comment |Â
I didn't know that multiplying an inequality by -1 changes the direction of the inequality sign. Thank you so much for pointing that out, Yves! Also, please, let me know if I should edit or do something with the post, because (I realise) it's really not a question about hypothesis rejection. Thank you.
– malasi
Jul 26 at 17:13
@malasi: I did'nt work this out with a multiplication but by changing members.
– Yves Daoust
Jul 26 at 17:44
what do you mean by "changing members?" I'm confused now.
– malasi
Jul 26 at 17:52
I got it, you simply added +2x to the inequality 5-2x ≥ 0. I had somehow gotten confused by the order of members. Thanks for the notice.
– malasi
Jul 26 at 20:13
I didn't know that multiplying an inequality by -1 changes the direction of the inequality sign. Thank you so much for pointing that out, Yves! Also, please, let me know if I should edit or do something with the post, because (I realise) it's really not a question about hypothesis rejection. Thank you.
– malasi
Jul 26 at 17:13
I didn't know that multiplying an inequality by -1 changes the direction of the inequality sign. Thank you so much for pointing that out, Yves! Also, please, let me know if I should edit or do something with the post, because (I realise) it's really not a question about hypothesis rejection. Thank you.
– malasi
Jul 26 at 17:13
@malasi: I did'nt work this out with a multiplication but by changing members.
– Yves Daoust
Jul 26 at 17:44
@malasi: I did'nt work this out with a multiplication but by changing members.
– Yves Daoust
Jul 26 at 17:44
what do you mean by "changing members?" I'm confused now.
– malasi
Jul 26 at 17:52
what do you mean by "changing members?" I'm confused now.
– malasi
Jul 26 at 17:52
I got it, you simply added +2x to the inequality 5-2x ≥ 0. I had somehow gotten confused by the order of members. Thanks for the notice.
– malasi
Jul 26 at 20:13
I got it, you simply added +2x to the inequality 5-2x ≥ 0. I had somehow gotten confused by the order of members. Thanks for the notice.
– malasi
Jul 26 at 20:13
add a comment |Â
up vote
1
down vote
In addition to what has been said, note that we can solve the equation by noting that
$$
|x|=aiff x=pm a
$$
In particular
$$
|5-2x|=8.25iff 5-2x=8.25 quad textor quad 5-2x=-8.25
$$
which is perhaps easier to solve.
answered Jul 26 at 15:50
Foobaz John
18k41245
add a comment |Â
up vote
1
down vote
In addition to what has been said, note that we can solve the equation by noting that
$$
|x|=aiff x=pm a
$$
In particular
$$
|5-2x|=8.25iff 5-2x=8.25 quad textor quad 5-2x=-8.25
$$
which is perhaps easier to solve.
answered Jul 26 at 15:50
Foobaz John
18k41245
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In addition to what has been said, note that we can solve the equation by noting that
$$
|x|=aiff x=pm a
$$
In particular
$$
|5-2x|=8.25iff 5-2x=8.25 quad textor quad 5-2x=-8.25
$$
which is perhaps easier to solve.
answered Jul 26 at 15:50
Foobaz John
18k41245
In addition to what has been said, note that we can solve the equation by noting that
$$
|x|=aiff x=pm a
$$
In particular
$$
|5-2x|=8.25iff 5-2x=8.25 quad textor quad 5-2x=-8.25
$$
which is perhaps easier to solve.
answered Jul 26 at 15:50
Foobaz John
18k41245
answered Jul 26 at 15:50
Foobaz John
18k41245
answered Jul 26 at 15:50
Foobaz John
18k41245
answered Jul 26 at 15:50
Foobaz John
18k41245
18k41245
add a comment |Â
add a comment |Â
up vote
0
down vote
A good way to solve this kind of equation is consider two cases
1) For $5-2xge 0 iff xle frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (5-2x) - 1/2 = 5 iff 20-8x-3=30 \iff 8x=-13 iff x=-frac138$$
that solution is acceptable since it is consistent to the assumption $xle frac25$.
2) For $5-2x< 0 iff x> frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (2x-5) - 1/2 = 5 iff 8x-20-3=30 \iff 8x=53 iff x=frac538$$
which is also acceptable.
As an example with some solution to reject refer to Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$.
answered Jul 26 at 15:45
gimusi
65k73583
add a comment |Â
up vote
0
down vote
A good way to solve this kind of equation is consider two cases
1) For $5-2xge 0 iff xle frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (5-2x) - 1/2 = 5 iff 20-8x-3=30 \iff 8x=-13 iff x=-frac138$$
that solution is acceptable since it is consistent to the assumption $xle frac25$.
2) For $5-2x< 0 iff x> frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (2x-5) - 1/2 = 5 iff 8x-20-3=30 \iff 8x=53 iff x=frac538$$
which is also acceptable.
As an example with some solution to reject refer to Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$.
answered Jul 26 at 15:45
gimusi
65k73583
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A good way to solve this kind of equation is consider two cases
1) For $5-2xge 0 iff xle frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (5-2x) - 1/2 = 5 iff 20-8x-3=30 \iff 8x=-13 iff x=-frac138$$
that solution is acceptable since it is consistent to the assumption $xle frac25$.
2) For $5-2x< 0 iff x> frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (2x-5) - 1/2 = 5 iff 8x-20-3=30 \iff 8x=53 iff x=frac538$$
which is also acceptable.
As an example with some solution to reject refer to Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$.
answered Jul 26 at 15:45
gimusi
65k73583
A good way to solve this kind of equation is consider two cases
1) For $5-2xge 0 iff xle frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (5-2x) - 1/2 = 5 iff 20-8x-3=30 \iff 8x=-13 iff x=-frac138$$
that solution is acceptable since it is consistent to the assumption $xle frac25$.
2) For $5-2x< 0 iff x> frac25$ we have
$$frac23 |5-2x| - 1/2 = 5 iff frac23 (2x-5) - 1/2 = 5 iff 8x-20-3=30 \iff 8x=53 iff x=frac538$$
which is also acceptable.
As an example with some solution to reject refer to Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$.
answered Jul 26 at 15:45
gimusi
65k73583
edited Jul 26 at 15:52
answered Jul 26 at 15:45
gimusi
65k73583
answered Jul 26 at 15:45
gimusi
65k73583
answered Jul 26 at 15:45
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
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1
Why do you say $|5-2x|=5-2x$ when $xge 2.5$?
– Lord Shark the Unknown
Jul 26 at 15:43