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# m25,#30

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Joined: 18 May 2008
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01 Dec 2008, 06:39
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Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

(A) $$\frac{1}{5}$$
(B) $$\frac{1}{4}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{3}{4}$$
(E) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can sum1 give a simpler explanation?
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Joined: 17 Jun 2008
Posts: 1547

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01 Dec 2008, 07:11
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Look at it this way.

Total part is 1. Out of this, x part was replaced.

Hence, (1-x)*50 (percent) + x*30 = 1*40
or, 50 - 20x = 40
or, 10 = 20x
or, x = 1/2....or, 50% of original solution was replaced.
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03 Feb 2009, 11:27
A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(A) 20% of all chips in the basket are green
(B) The ratio of the number of red chips to the number of green chips is 4:1

I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.

Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?
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03 Feb 2009, 12:27
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jairus wrote:
A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(A) 20% of all chips in the basket are green
(B) The ratio of the number of red chips to the number of green chips is 4:1

I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.

Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?

A: you need the total number of chips. ok, 20% of all chips in the basket are green.
total = 100
green = 20
red = 80
the prob = (20/100) x (19/99) = 19/(5x99)

suppose the total = 20
green = 4
red = 16
the prob = (4/20) x (3/19) = 3/(5x19)

see how the prob differes with the number of chips in the bowl.

(B) again, its same with B too even if the ration of red to green is 4:1. try with some number.

togather also not clear.

Therefore, it is E.
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17 Dec 2009, 23:55
ritula wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?

(C) 2008 GMAT Club - m25#30

Can sum1 give a simpler explanation?

I understand the alegbraic approach here but can this question be solved using more basic math principles?

If a solution goes from 50% to 40% in a mixture with 30% of some proportion, doesnt logic dictate that the mixture be 50/50?

i understand this approach doesnt work with many other mixture problems, but in this case where equal amounts are used and it is asking about proportions and not the overall amounts used, wouldnt it work? thoughts?
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14 Sep 2010, 17:06
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ritula wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

(A) $$\frac{1}{5}$$
(B) $$\frac{1}{4}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{3}{4}$$
(E) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

Can sum1 give a simpler explanation?

Number and Answer back substitution works quite well for these kind of problems.
First convert % to a number(100 usually) and also choose a quantity, like ml.
Say you had 100ml of the solution with 50ml acid and 50ml water, and you want to end up with 40ml of acid(40%)

Pick Choice A - 1/5
1/5 solution = 1/5(100ml) = 20ml.
Because you remove a a 50/50 solution, you remove 10ml acid and 10ml water. Because you add a 30/70 solution, it's 6ml acid(30% of 20ml) and 14ml(70% of 20ml) water.
So you have 50ml(original 50/50) - 10ml(50/50) + 6ml(30/70) = 46ml acid - this is different from the 40%

You also know that you didn't remove enough of the original solution, so skip B and jump to choice C. (I usually start with choice C anyway for answer substitution)

Pick Choice A - 1/2
1/2 solution = 1/2(100ml) = 50ml.
Because you remove a a 50/50 solution, you remove 25ml acid and 25ml water. Because you add a 30/70 solution, it's 15ml acid(30% of 50ml) and 35ml water(70% of 20ml).
So you have 50ml(original 50/50) - 25ml(50/50) + 15ml(30/70) = 40ml acid - this is 40%, and is hence the correct answer.
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14 Sep 2010, 17:10
jairus wrote:
A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(A) 20% of all chips in the basket are green
(B) The ratio of the number of red chips to the number of green chips is 4:1

I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.

Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?

You came up with the answer 0 for 4:1.
Now if you choose red:green=8:2 (which is equivalent to the ratio 4:1), you get an answer(2/10*1/9) which is different from 0. So its E
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15 Sep 2010, 04:18
mustafaj wrote:
ritula wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?

(C) 2008 GMAT Club - m25#30

Can sum1 give a simpler explanation?

I understand the alegbraic approach here but can this question be solved using more basic math principles?

If a solution goes from 50% to 40% in a mixture with 30% of some proportion, doesnt logic dictate that the mixture be 50/50?

i understand this approach doesnt work with many other mixture problems, but in this case where equal amounts are used and it is asking about proportions and not the overall amounts used, wouldnt it work? thoughts?

In this particular question, before replacement, solution is 50% and after replacement it is 40%. Since we are given that the added solution is 30%, we can think of basic average formula which is (50+30)/2 = 40. Hence both 50% and 30% solutions should be same and is equal to half.
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17 Sep 2010, 13:54
A bowl contains chips of red and green color and no chips of any other color. If two chips are drawn from the basket at random without replacement, what is the probability that both chips will be green?

(A) 20% of all chips in the basket are green
(B) The ratio of the number of red chips to the number of green chips is 4:1

I don't understand the explanation: First of all, both statements give the same information (80% : 20% = 4 : 1). Second, this information is not sufficient; the answer will depend on the total number of chips in the basket. As an example, consider the case when the basket contains 5 chips; the answer is 0.

Why is it not D? There can't be 5 chips cause we can't have fractioned chips, can we? If we had, we would get the solution anyway because we have the ratio blule to red chips, wouldn't we?

Addressing your problem - your intent is right, however what you have to realise is that you are only provided with the ratios in the quesion and not specefic values. It is easy assume that 20 out of 100 or 2 out of 10 chips are green, which is not the case here.

Option 1 says - 20% of chips are green. Now this info is clearly not enough as we don't know total number of chips. So we can't say whether number of green chips are 20 or 200 or 200000. So choice 1 is clearly insufficient.

Now if you get my point above, you'd realise that Option 2 is just a re-wording of the option 1, i.e. if the ratio is 4:1, then green chips are certainly 20% of the total.

I hope you understand why D is wrong. D is incorrect since we do not have values and only have ratios. If the question had asked us to pick just one chip, D wld have been correct.

Cheers!
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Joined: 11 May 2011
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15 Sep 2011, 07:16
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krishnasty wrote:
Although i got the ans through conventional method, can somebody please solve this question through amalgation method used in mixture problems?

This one is cake. Answer C.
Can be solved on 30 sec. It is just becasue data is very simple.

50% solution of acid and 30% solution of acid = 40% solution of acid -- Straight average of both acid is required. Hence, 50-50 of both. Answer C. Hope it helps.

Cheers!

P.S. - This is applicable for all the questions in which final concentration is becoming average concentration of 2 solutions which are getting mixed.
10% solution of acid and 30% solution of acid = 20% solution of acid
20% solution of acid and 40% solution of acid = 30% solution of acid
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17 Sep 2011, 02:22
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Yup its C. Since (50+30) /2= 40 => The resulting solution should have half of each of the solutions.
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22 Mar 2012, 20:23
you can easily solve this question by just seeing that 40% is also in the mid of the two numbers and to be in the middle you must have an equal amount of both.
Caution though that this will only work in this situation where it is perfectly in the middle of the two numbers... it would be harder to tell if the numbers were larger and you couldnt see the middle number of the two.
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26 Mar 2012, 14:19
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I think you can look at this conceptually to safe a lot of time.
It's clearly a weighted average question.

30%-----40%-----50%

40% is perfectly in the middle, therefore you need 1 part 30% solution to each part 50% solution. The mixture will contain 1/2 30% mix and 1/2 50% mix.
You have removed 1/2 of the original 50% solution and replaced it with 30% solution.

The tricky thing here is that if you didn't see this and you were looking to eliminate answers using the method described in Mgmat where you look for fractional values that add to 1, you would eliminate answer (C) for not having a match.

Done!
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17 Sep 2012, 05:14
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Alligative Method is such a time-saving method.

Here we go:

30%-----40%-----50%
Distances 1:1

This means that both contribute an equal amount of the final solution.

Straight C: 1/2

Cheers,
Der alte Fritz.
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26 Sep 2012, 11:49
I solved this question by following method without being sure if it was true, but it gave a correct answer and took some times ...

0.5x - a0.5x + a0.3x = 0.4x (1)

a is the same amount of solutions replaced

from (1) we have: a0.2x = 0.1x
thus, a = 1/2

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26 Sep 2012, 12:07
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Expert's post
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$

Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

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16 Sep 2013, 23:18
Bunuel wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$

Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

I wonder if anybody was confused with the word "replaced" at all? I was trying to come up with some different math only because of that word. Had the question said the solution was "added" to the original it would make it as simple as Bunuel put it. So it seems to me that "replace" = "add"in this context...disappointed how you can know the solution but wording messes up with you to the point you begin doubting what you can...

Bunuel - are there instances in mixtures when "replace" is not equal to "add"?
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17 Sep 2013, 00:58
obs23 wrote:
Bunuel wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

A. $$\frac{1}{5}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
E. $$\frac{4}{5}$$

Since the concentration of acid in resulting solution (40%) is the average of 50% and 30% then exactly half of the original solution was replaced.

Alternately you can do the following, say $$x$$ part of the original solution was replaced then: $$(1-x)*0.5+x*0.3=1*0.4$$ --> $$x=\frac{1}{2}$$.

I wonder if anybody was confused with the word "replaced" at all? I was trying to come up with some different math only because of that word. Had the question said the solution was "added" to the original it would make it as simple as Bunuel put it. So it seems to me that "replace" = "add"in this context...disappointed how you can know the solution but wording messes up with you to the point you begin doubting what you can...

Bunuel - are there instances in mixtures when "replace" is not equal to "add"?

Replacing and adding are not the same. Here we are told that "some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid".
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17 Sep 2013, 03:27
Quote:
Replacing and adding are not the same. Here we are told that "some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid".
[/quote]

But then, mathematically, we seem to treat it identically? My mathematical equation $$0.5x + 0.3(1-x)=0.4$$ was identical to yours right away, but then I started questioning, thinking since we are "replacing" we could not express this relationship the same way, in the weighted average manner...I am not sure what I am missing, but it does not feel right
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17 Sep 2013, 03:44
obs23 wrote:
Quote:
Replacing and adding are not the same. Here we are told that "some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid".

But then, mathematically, we seem to treat it identically? My mathematical equation $$0.5x + 0.3(1-x)=0.4$$ was identical to yours right away, but then I started questioning, thinking since we are "replacing" we could not express this relationship the same way, in the weighted average manner...I am not sure what I am missing, but it does not feel right [/quote]

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# m25,#30

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text:
# Weierstrass $\wp$-Function Addition Property

Consider the function

$$\det\left( \begin{array}{ccccc} &1 &\wp(z) &\wp'(z) \\ &1 &\wp(w) &\wp'(w) \\ &1 &\wp(-z-w) &\wp'(-z-w) \end{array} \right)=f(z)$$

I'm trying to prove that it has at least $6$ distinct zeroes if $w\notin\frac{1}{3}\Lambda$ and at least $5$ zeroes with multiplicity otherwise. Then by a degree argument $f$ is obviously identically zero. I've got no idea how to show this though! Obviously $f$ has a zero at $z=w$. Apart from that I know the zeroes of $\wp'$ but they seem to be of no help. It seems like I'm meant to try $z=3w$ but this doesn't work! A hint would be greatly appreciated!

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See here for a nice outline: math.stackexchange.com/q/121662/5363 –  t.b. Jun 1 '12 at 11:16
I've read that and don't really understand it at all! I think that this question is meant to be solved much more simply, since we haven't really talked about any of the theory in the course. –  Edward Hughes Jun 1 '12 at 11:18
You could simply use the expansions and verify that the principal part vanishes, then apply Liouville (note that $f$ is $\Lambda$-periodic by definition). This makes a straightforward if somewhat messy calculation. –  t.b. Jun 1 '12 at 11:20
Maybe I'm being stupid but I don't see how that tells me how many zeroes I have! –  Edward Hughes Jun 1 '12 at 11:21
Okay have found two more, namely $z=-2w$, $z=-\frac{1}{2}w$ but these put different constraints on $w$ to the ones I've been given! –  Edward Hughes Jun 1 '12 at 11:22

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text:
The knot whose complement is the Hantzsche-Wendt manifold

Can someone identify the knot whose complement in $\mathbb{S}^{3}$ will produce the Hantzsche-Wendt manifold?

Thanks

-
As far as I know, the HW manifold is closed (it is the flat 3-manifold that Seifert-fibers over the projective plane). –  Bruno Martelli Jan 26 '13 at 6:56
What is the definition of this manifold? There's not much that's googlable. –  Ryan Budney Jan 26 '13 at 7:47
The HW manifold is the orbit space $\mathbb{R}^{3} / <x , y , z |xy^{2}x^{-1}y^{2} = y x^{2} y^{-1} x^{2} = 1 , z = xy >$ where $X = diag[1 , -1 , 1] , Y = diag[-1, 1 , -1 ]$ and $Z = diag[-1,-1,1]$ and $x = (\frac{1}{2} e_{1} , X), y = ( \frac{1}{2} (e_{2} - e_{3} ) , Y )$ and $z=(\frac{1}{2}(e_{1} - e_{2} + e_{3} ) , Z)$, this group is a subgroup of the group of affine motions of $3$-space. –  jf9rj4o Jan 26 '13 at 13:04
@jf9: Does "produce" mean Dehn suegery? Otherwise, Bruno answered your question. –  Misha Jan 26 '13 at 18:43
add comment

2 Answers

Much enlightenment (though not an explicit answer to the question) can be gleaned from Bruno Zimmermann's paper "On the Hantzsche-Wendt manifold".

-
add comment

Supposing the question is, "Can the Hantsche-Wendt manifold be realized as a cyclic branched cover of a knot in the 3-sphere", then the answer is yes: it is a 3-fold cyclic branched cover along the figure-eight knot. Incidentally, it is also a 2-fold branched cover along the Borromean rings.

On the other hand, if the question is, "Can the Hantsche-Wendt manifold $M$ be realized by performing Dehn surgery along a knot in the 3-sphere" (as Misha seems to suggest), then the answer is no, since $H_1(M,\mathbb{Z})=\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is not a cyclic group.

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As shown in the reference I gave. –  Igor Rivin Jan 27 '13 at 2:08
Sorry, I saw your post after I finished writing my answer. And yes, that's the right reference. –  Alex Suciu Jan 27 '13 at 2:51
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text:
# A non locally compact group of finite topological dimension?

Is there a topological group which is Hausdorff, first countable, locally connected and has finite topological dimension, yet fails to be locally compact?

• Is there a topological space satisfying these properties? Feb 11 '16 at 19:31
• @YonatanHarpaz: Remove from the plane those points which have rational coordinates. Feb 11 '16 at 19:57
• @YonatanHarpaz for spaces, in dimension 1 there are examples as well. Just take the 1-skeleton of a graph not of finite degree, endowed with either the direct limit topology (of its finite subgraphs), or with its distance topology.
– YCor
Feb 11 '16 at 22:05

The Gleason-Montgomery theorem on the local compactness of locally path-connected finite-dimensional topological groups was generalized by Banakh and Zdomskyy (http://topology.auburn.edu/tp/reprints/v36/tp36027.pdf) who proved that a topological group $G$ is locally compact if it is compactly finite-dimensional and locally continuum-connected.

On the other hand, for every $n\ge 2$ the Euclidean space $\mathbb R^n$ contains an additive subgroup $G$, which is connected, locally connected, and not Borel (so, not locally compact).

To construct such a group $G$, consider the family $\mathcal B$ of all uncountable Borel subsets of $\mathbb R^n$ and observe that it has cardinality of continuum. So, $\mathcal B$ can be enumerated by ordinals $<\mathfrak c$ as $(B_\alpha)_{\alpha<\mathfrak c}$. Fix any non-zero point $p$ in $\mathbb R^n$ and by transfinite induction, for every ordinal $\alpha<\mathfrak c$ choose a point $x_\alpha\in B_\alpha$ so that the subgroup $G_\alpha$ generated by the set $\{x_\beta\}_{\beta\le\alpha}$ does not contain the point $p$. Such a choice is always possible since $B_\alpha$ has cardinality of continuus and $x_\alpha$ should be chosen to avoid the subset $\{\frac1n(p-g):n\in\mathbb Z\setminus\{0\},\;g\in \bigcup_{\beta<\alpha}G_{\beta}\}$ of cardinality $<\mathfrak c$. After completing the inductive construction, consider the additive subgroup $G$ of $\mathbb R^n$, generated by the set $\{x_\alpha\}_{\alpha<\mathfrak c}$. By construction, this subgroup $G$ does not contain the point $p$ and meets every uncountable Borel subset $B$ of $\mathbb R^n$. In particular, it meets the uncountable Borel set $B-p$, which implies that $p+G\subset \mathbb R^n\setminus G$ meets $B$. This means that $G$ is a Bernstein set in $\mathbb R^n$.

It remains to prove that every Bernstein set $G$ in $\mathbb R^n$, $n\ge 2$, is connected and locally connected. This will follow as soon as we check that for every open subset $U\subset\mathbb R^n$, homeomorphic to $\mathbb R^n$ the intersection $U\cap G$ is connected. To derive a contradiction, assume that $U\cap G$ is disconnected and find two disjoint non-empty open sets $U_1,U_2\subset U$ such that $U\cap G=(U_1\cap G)\cup(U_2\cap G)$. It follows that $F=U\setminus (U_1\cup U_2)$ is a closed subset of $U$ separating $U$, which is disjoint with the Bernstein set $G$. So, $F$ is at most countable and hence $\dim(F)=0$. But this contradicts Mazurkiewicz Theorem (saying that $\mathbb R^n$ cannot be separated by a closed subset of dimension $<n-1$).

This is not quite an answer to the question, but it may be of interest: this paper (see Theorem 1) shows that a locally connected, arcwise-connected finite-dimensional topological group is a Lie group.

Of course we also know that a path-connected Hausdorff space is arcwise connected (see the literature on Peano spaces and the Hahn-Mazurkiewicz theorem, as mentioned for instance here), so we can say that a path-connected locally connected finite-dimensional $T_0$ topological group is a Lie group. That gets you pretty close, if not a cigar.

• Theorem 3 here: pnas.org/content/36/11/663, says that any locally arcwise-connected group of finite dimension is locally compact. On the other hand I don´t know any metrizable locally connected topological space (group or otherwise) which is not locally path-connected. Feb 11 '16 at 20:03
• Of course :P ... and locally compact? Feb 11 '16 at 20:22


text:
Collective-Mode Enhanced Matter-Wave Optics.

@article{Deppner2021CollectiveModeEM,
title={Collective-Mode Enhanced Matter-Wave Optics.},
author={Christian Deppner and Waldemar Herr and Merle Cornelius and Peter Stromberger and Tammo Sternke and Christoph Grzeschik and Alexander Grote and Jan Rudolph and Sven Herrmann and Markus C. Krutzik and Andr'e Wenzlawski and Robin Corgier and Eric Charron and D. Gu{\'e}ry-Odelin and Naceur Gaaloul and Claus L{\"a}mmerzahl and Achim Peters and Patrick Windpassinger and Ernst Maria Rasel},
journal={Physical review letters},
year={2021},
volume={127 10},
pages={
100401
}
}
• Published 3 September 2021
• Physics
• Physical review letters
In contrast to light, matter-wave optics of quantum gases deals with interactions even in free space and for ensembles comprising millions of atoms. We exploit these interactions in a quantum degenerate gas as an adjustable lens for coherent atom optics. By combining an interaction-driven quadrupole-mode excitation of a Bose-Einstein condensate (BEC) with a magnetic lens, we form a time-domain matter-wave lens system. The focus is tuned by the strength of the lensing potential and the…
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A high-flux BEC source for mobile atom interferometers

• Physics
• 2015
Quantum sensors based on coherent matter-waves are precise measurement devices whose ultimate accuracy is achieved with Bose–Einstein condensates (BECs) in extended free fall. This is ideally

Design of a dual species atom interferometer for space

• Physics
• 2014
Atom interferometers have a multitude of proposed applications in space including precise measurements of the Earth’s gravitational field, in navigation & ranging, and in fundamental physics such as

Bose-Einstein Condensation in Microgravity

• Physics
Science
• 2010
The preparation and observation of a Bose-Einstein condensate during free fall in a 146-meter-tall evacuated drop tower is reported and represents a promising source for matter-wave interferometry to test the universality of free fall with quantum matter.

Interacting quantum mixtures for precision atom interferometry

• Physics
New Journal of Physics
• 2020
We present a source engineering concept for a binary quantum mixture suitable as input for differential, precision atom interferometry with drift times of several seconds. To solve the non-linear

Fast manipulation of Bose–Einstein condensates with an atom chip

• Physics
• 2017
We present a detailed theoretical analysis of the implementation of shortcut-to-adiabaticity protocols for the fast transport of neutral atoms with atom chips. The objective is to engineer transport

STE-QUEST—test of the universality of free fall using cold atom interferometry

• Physics
• 2013
The theory of general relativity describes macroscopic phenomena driven by the influence of gravity while quantum mechanics brilliantly accounts for microscopic effects. Despite their tremendous

Improving the accuracy of atom interferometers with ultracold sources

• Physics
New Journal of Physics
• 2018
We report on the implementation of ultracold atoms as a source in a state of the art atom gravimeter. We perform gravity measurements with 10 nm s−2 statistical uncertainties in a so-far largely

The Bose-Einstein Condensate and Cold Atom Laboratory

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EPJ Quantum Technology
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Microgravity eases several constraints limiting experiments with ultracold and condensed atoms on ground. It enables extended times of flight without suspension and eliminates the gravitational sag

Inertial sensing with quantum gases: a comparative performance study of condensed versus thermal sources for atom interferometry

• Physics
• 2021
Abstract Quantum sensors based on light pulse atom interferometers allow for measurements of inertial and electromagnetic forces such as the accurate determination of fundamental constants as the

ELGAR—a European Laboratory for Gravitation and Atom-interferometric Research

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Classical and Quantum Gravity
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Gravitational waves (GWs) were observed for the first time in 2015, one century after Einstein predicted their existence. There is now growing interest to extend the detection bandwidth to low


text:
Lesson plan

# Lesson 26: Testing Points to Solve Inequalities

teaches Common Core State Standards MP7 http://corestandards.org/Math/Practice/MP7
teaches Common Core State Standards MP3 http://corestandards.org/Math/Practice/MP3
teaches Common Core State Standards 8.EE.C.7.b http://corestandards.org/Math/Content/8/EE/C/7/b
teaches Common Core State Standards 8.EE.C.7 http://corestandards.org/Math/Content/8/EE/C/7

# Lesson 26: Testing Points to Solve Inequalities

The mathematical purpose of this lesson is to give students a chance to name and analyze common incomplete strategies and concepts when working with inequalities. Students engage in error analysis on the topic of testing points around a given boundary to determine which half of the real number line contains solutions to an inequality, and on the the topic of solving equations.

Error analysis around solving equations helps students continue to hone the foundational skill of solving equations, which is useful for finding the boundary point of the solutions of a linear inequality. Exploring the reasoning behind testing points helps students make strategic choices as they solve inequalities.

In the supported Algebra 1 lesson, students are using their understanding of systems of inequalities with two variables to create a mathematical model for designing trail mix.

Students have the opportunity to construct viable arguments and critique the reasoning of others (MP3) as they engage in error analysis.

Lesson overview

• 26.1 Warm-up: Math Talk: Solving Equations (10 minutes)
• 26.2 Activity: Character Chat (20 minutes)
• 26.3 Activity: Error! (15 minutes)

Learning goals:

• Analyze common errors when solving linear equations.
• Understand the purpose, when solving inequalities, of testing points on either side of the solution to the related equation.

Learning goals (student facing):

• Let’s critique some peoples’ reasoning.

Standards:

• This lesson builds towards the standards: CCSS.HSA-REI.B.3MS.A-REI.3MO.A1.CED.A.1

IM Algebra 1, Geometry, Algebra 2 is copyright 2019 Illustrative Mathematics and licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0).

The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics.


text:
A primer on universal function approximation with deep learning (in Torch and R)

Arthur C. Clarke famously stated that “any sufficiently advanced technology is indistinguishable from magic.” No current technology embodies this statement more than neural networks and deep learning. And like any good magic it not only dazzles and inspires but also puts fear into people’s hearts. This primer sheds some light on how neural networks work, hopefully adding to the wonder while reducing the fear.

One known property of artificial neural networks (ANNs) is that they are universal function approximators. This means that any mathematical function can be represented by a neural network. Let’s see how this works.

Function Approximation

Universal function approximation may sound like a powerful spell recently discovered. In fact, function approximation has been around for centuries. Common techniques include the Taylor series and the Fourier series approximations. Recall that given enough terms, a Taylor series can approximate any function to a certain level of precision about a given point, while a Fourier series can approximate any periodic function. By definition, these series are linear combinations of multiple terms. The same is true of neural networks, where output values are linear combinations of their input. The difference is that neural networks can approximate a function over a set of points.

To illustrate, consider the function $z = 2x^2 - 3y^2 + 1$. This function describes a saddle, as shown in Figure 1. This function is clearly nonlinear and cannot be accurately approximated by a plane (such as a linear regression). Can NNs do better?

Figure 1. The saddle function 2x^2 – 3y^2 + 1.

For sake of argument, let’s attempt to approximate this function with a $2 \times 1$ network anyway. In its most basic form, a one layer neural network with $latex$n input nodes and one output node is described by $y = Ax + b$, where $x$ is the input, $b$ is the bias and $A$ is the weight matrix. In this case, since $y$ is a scalar, $A$ is simply a vector. In Torch, this network can be described by

require "nn"model = nn.Sequential()model:add(nn.Linear(2,1))

After training, we can apply the training set to the model to see what the neural network thinks $z$ looks like. Not surprisingly it’s a plane, which is far from what we want.

Let’s bring in the cavalry and add a hidden layer to create a $2 \times 10 \times 1$ network. Surely with a hidden layer (and more parameters) the network can do a better job of approximating $z$. In Lua, we say

require "nn"model = nn.Sequential()model:add(nn.Linear(2,10))model:add(nn.Linear(10,1))

Alas, it isn’t so. The result is still a plane, albeit slightly rotated.

Before you lose faith in artificial neural networks, let’s understand what’s happening. In the single layer network with one output node, we said $A$ was a vector. In this case $A$ is a proper matrix since the output is a hidden layer with 10 nodes. So $y$ is a vector of size 10. In the output layer we have a new weight matrix and bias term applied to $y$. Mathematically, the second layer is simply applying a second linear transformation. In others words, we still get a plane, since all terms are linear! To move beyond linear functions, we need to inject nonlinearity in the model.

Activation Functions

Linearity is an exceptionally useful property, but in terms of functions, it’s not particularly interesting. The nonlinear world is far more mysterious and exciting, and so it is with neural networks. Nonlinearity is injected into neural networks through the use of an activation (aka transfer) function. This function $g$ is applied to the linear component: $y = g(Ax + b)$. The choice for $g$ ranges from the sigmoid to tanh to the default rectified linear unit (ReLU).

require "nn"model = nn.Sequential()model:add(nn.Linear(2,10))model:add(nn.Tanh())

The result of this model is starting to look like the target function. So we’re on the right track.

However, the result is rather blocky. This suggests that there are not enough parameters to properly capture the nonlinearity. By increasing the hidden layer to 40 nodes, we can smooth this out further.

Development Notes

I used Torch to generate the training set and the neural networks. To analyze the results I used standard R.

Training the network requires a test dataset. I generated one (called trainset below) using Lua that conformed to the Torch spec for row-major datasets. To train the model, I used the SmoothL1Criterion coupled with stochastic gradient descent. Good results require fiddling with the learningRate and the learningRateDecay on the trainer. (I don’t include these parameters as this is related to an exercise for my students.)

criterion = nn.SmoothL1Criterion()trainer = nn.StochasticGradient(model, criterion)trainer:train(trainset)

To render the plots, I evaluate the trained model against the complete training set and write out a CSV. Then I load into R and use the scatterplot3D package to render the surface.

Conclusion

Function approximation is a powerful capability of neural networks. In theory any function can be approximated, but constructing such a network is not so trivial. One of the key lessons with neural networks is that you cannot blindly create networks and expect them to yield something useful. Not only does it take patience, but it takes an understanding and appreciation of the theory to lead you down the correct path.

Brian Lee Yung Rowe is an Adjunct Professor at the CUNY MS Data Analytics program and also the founder of Pez.AI, a chat-based data analysis and computing platform with a conversational AI interface. This article is reformatted for this blog. View the original on Cartesian Faith.


text:
# Best way of transforming random variable

I am working with the following code:

\[ScriptCapitalD] =
TransformedDistribution[u - Log[u],
u \[Distributed] UniformDistribution[{1, 2}]]
Plot[PDF[\[ScriptCapitalD], x], {x, 1, 2}, Filling -> Axis]

Basically, I want to transform a uniform random variable, $$u$$ into $$u-\log{u}$$, and plot the pdf of the new random variable. However, Mathematica seems to take a long time using the function TransformedDistribution and moreover, I am not being able to retrieve a Plot, so perhaps the function is not working well under my transformation.

Do you have any suggestions on how to obtain a solution for this?

• Cross-posted at math.stackexchange.com/questions/3625412/….
– JimB
Apr 15 '20 at 18:03
• I don't understand why @m0nhawk deleted his/her correct answer. Apr 15 '20 at 18:25
• @user64494 $U$ and $\log U$ are not independent as assumed in that answer.
– JimB
Apr 15 '20 at 18:34
• @JimB: I've demonstrated how to correct it, considering two IID. Apr 15 '20 at 18:48
• Your question on the Math StackExchange site was (essentially) about $u-\log u$. Here did you mean $u+\log u$ or $u-\log u$ ?
– JimB
Apr 15 '20 at 19:58

It might just be with the Plot command that is giving you trouble.

dist = TransformedDistribution[ u - Log[u], u \[Distributed] UniformDistribution[{1, 2}]];
pdf = PDF[dist, z] // FullSimplify

results in

To plot that one can use

Plot[Evaluate[pdf[[1, 1, 1]]], {z, 1, 2 - Log[2]}, PlotRange -> {Automatic, {0, Automatic}}]

As a check:

zz = RandomVariate[dist, 100000];
Show[Histogram[zz, 100, "PDF"],
Plot[Evaluate[pdf[[1, 1, 1]]], {z, 1, 2 - Log[2]}, PlotRange -> {Automatic, {0, 30}}]]

The resulting Piecewise function from PDF[dist, z] has two pieces that are "identities" that are true when 1 <= z <= 2 - Log[2] but when Plot evaluates the function with machine precision numbers, things go weird. So here is a cleaned up version of the resulting probability density function that plays well with Plot:

dist = TransformedDistribution[ u - Log[u], u \[Distributed] UniformDistribution[{1, 2}]];
pdf = PDF[dist, z] // TrigToExp // FunctionExpand;
pdf = Piecewise[{{pdf[[1, 1, 1]], 1 <= z <= 2 - Log[2]}}, 0]

Plot[pdf, {z, 0.9, 3/2}, PlotStyle -> Thickness[0.01], PlotRangeClipping -> None]

• Something to adjust in view of the result of Plot[Evaluate[pdf[[1, 1, 1]]], {z, 0, 4}, PlotRange -> {Automatic, {0, Automatic}}] . Apr 15 '20 at 20:30
• Also Plot[Evaluate[pdf], {z, 0, 4}, PlotRange -> {Automatic, {0, Automatic}}] produces an error communication and an empty plot. Apr 15 '20 at 20:43
• @user64494 Thanks, that makes it better: I'll modify the PlotRange as you suggest. On your other comment: that happens because of the conditions of the pdf are not always true when using machine precision numbers that Plot uses. I'll put in a fix for that.
– JimB
Apr 15 '20 at 20:53
• Looks good, thanks! Is there a way to introduce a parameter when working with """RandomVariate?""" Say """dist[a_] = TransformedDistribution[a*u - Log[u], u [Distributed] UniformDistribution[{1, 2}]]; zz[a_] = RandomVariate[dist[a], 100000];"""? And then use """Manipulate""" over the parameter $a$? Apr 15 '20 at 20:53
• @user64494 I'm sure someone could supply the steps. And it was not just made by hand; my small brain helped a bit.
– JimB
Apr 15 '20 at 21:26


text:
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## Question

###### ConsincDndanWhe holnehai Is thc Vnltact CrUss th ( coucorInttart J03t brerJolljin 42JL0 R01940 !Tnes 0 20Atcrtheynon /5 doscJlora time wha: Ehe neror

Consinc Dndan Whe holne hai Is thc Vnltact CrUss th ( coucor Inttart J03t brer Jolljin 42 JL0 R 01940 ! Tnes 0 20 Atcrtheynon /5 doscJ lora time wha: Ehe neror

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-- 0.020782--


text:
# Get RGB value from realsense d435i pointcloud in simulation

Hi everyone,

i got a gazebo simulation running with a realsense d435i camera in it. All the proper ROS-topics are published and i can get the recorded pointcloud from the topic "/camera/depth/color/points" without a problem. My Problem is that I also need to get the RGB values of the points and I am struggling with that. I used this kind of script for it:

import sensor_msgs.point_cloud2 as pc2
import ctypes
import struct

#subscribe to pointcloud topic define the function that does the parsing
def PointCloud(self, PointCloud2):
self.lock.acquire()
int_data = list(gen)
for x in int_data:
test = x[3]
# cast float32 to int so that bitwise operations are possible
s = struct.pack('>f' ,test)
i = struct.unpack('>l',s)[0]
# you can get back the float value by the inverse operations
pack = ctypes.c_uint32(i).value
r = (pack & 0x00FF0000)>> 16
g = (pack & 0x0000FF00)>> 8
b = (pack & 0x0000FF00)
print r,g,b # prints r,g,b values in the 0-255 range


text:
Finding induced isomorphism from $\DeclareMathOperator{\Hom}{Hom} \Hom_F(U\oplus U_2, V)\to \Hom_f(U_, V)\oplus \Hom_f(U_2, V)$

I am seeing this kind of question a lot in my class, but each time fail to understand how to think about the question. Why is there an isomorphism between sets of homomorphisms and what is it explicitly?

Let $R$ be a ring and $U_1, U_2, V$ be $R$-modules. First, recall that we have canonical injections into the coproduct: \begin{align*} \iota_1: U_1 &\hookrightarrow U_1 \oplus U_2\\ u_1 &\mapsto (u_1, 0)\\ \iota_2: U_2 &\hookrightarrow U_1 \oplus U_2\\ u_2 &\mapsto (0, u_2) \, . \end{align*}

Given $R$-linear maps $\varphi_1: U_1 \to V$ and $\varphi_2: U_2 \to V$, the universal property of the coproduct states that there is a unique $R$-linear map $\varphi = \varphi_1 \oplus \varphi_2: U_1 \oplus U_2 \to V$ such that the following diagram commutes.

$\hskip1.9in$

Commutativity of the diagram implies that $(\varphi_1 \oplus \varphi_2)(u_1, u_2) = \varphi_1(u_1) + \varphi_2(u_2)$.

Conversely, given an $R$-linear map $\varphi: U_1 \oplus U_2 \to V$, pre-composition by the injections yields maps $U_1 \to V$ and $U_2 \to V$. Explicitly, let $\varphi_1 = \varphi \circ \iota_1 : U_1 \to V$ and $\varphi_2 = \varphi \circ \iota_2 : U_2 \to V$.

Thus the isomorphism you seek is \begin{align*} \hom_R(U_1, V) \oplus \hom_R(U_2,V) &\longrightarrow \hom_R(U_1 \oplus U_2, V)\\ (\varphi_1, \varphi_2) &\longmapsto \varphi_1 \oplus \varphi_2\\ \end{align*} with inverse \begin{align*} \hom_R(U_1 \oplus U_2, V) &\longrightarrow \hom_R(U_1, V) \oplus \hom_R(U_2,V)\\ \varphi &\longmapsto (\varphi \circ \iota_1, \varphi \circ \iota_2) \, . \end{align*}

• Diagram, you say? ;) (not required) – BrianO Oct 30 '15 at 18:44
• Working on it! :) – Viktor Vaughn Oct 30 '15 at 18:45
• Good show! (and good answer too) – BrianO Oct 30 '15 at 19:28


text:
Can URL encoding be disabled? [closed]

Hi! Is there a way to disable URL encoding for non-ASCII (UTF-8) characters? Once saving the HTML file the links turn into URL encoding format and IE is unable to read them. Thnx for help!

Example:

Prior editing and saving in LibreOffice:

  <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">

<meta http-equiv="content-type" content="text/html; charset=windows-1250">

Íii

After editing and saving in LibreOffice:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">

<meta http-equiv="content-type" content="text/html; charset=windows-1250">
<title></title>
<meta name="created" content="0;0">
<meta name="changed" content="20140311;164828687000000">

edit retag reopen merge delete

Closed for the following reason the question is answered, right answer was accepted by Alex Kemp close date 2016-02-20 07:30:30.921183

Sort by » oldest newest most voted

The underlying problem seems to be charset=windows-1250, IE seems to stumble over the URL being encoded in UTF-8 while the meta tag says it would be windows-1250. If you could make that use charset=utf-8 instead things might work.

more

Well the charset change doesn't help. It is because the IE doesn't support utf-8 format of url at all. Check here http://support.microsoft.com/kb/941052

( 2014-03-12 15:13:24 +0200 )edit

The original example is using Windows 1250 encoding (not UTF-8) i.e., Í is encoded as 013a (hex) / 229 (decimal), whereas in UTF-8 this would be U+00cd. LO is converting the Windows 1250 character to percent encoding, as per the W3C (RFC standard) recomendation. @erAck is correct in that with charset=utf-8 I can save the original HTML under v4.1.4.2 and obtain Íii which is valid UTF-8 encoding. If your host operating system is set to use Windows 1250 encoding you will have a larger issue.

( 2014-03-13 00:27:48 +0200 )edit

Well strange. I of course changed the charset and tried but it is not working. I filled a bug report so we will see. https://bugs.freedesktop.org/show_bug.cgi?id=76080

( 2014-03-13 07:59:32 +0200 )edit

According to Wikipedia Percent encoding a.k.a. URL encoding is supposed to be done for any character in a URI not defined by:

RFC 3986 section 2.2, Reserved Characters:

! * ' ( ) ; : @ & = + \$ , / ? # [ ]

RFC 3986 section 2.3, Unreserved Characters:

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

a b c d e f g h i j k l m n o p q r s t u v w x y z

0 1 2 3 4 5 6 7 8 9 - _ . ~

... in other words, most non-ASCII characters should be percent encoded.

more

From Microsoft IE support page: "To represent a non-US-ASCII character, you must use that non-US-ASCII character directly in the encoding of the document in which you write the URI."

Check more: http://support.microsoft.com/kb/941052

So using UTF-8-encoded characters is definetly not a good solution while big portion of users still uses IE.

( 2014-03-12 15:23:52 +0200 )edit

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text:
# Ring homomorphism and units homomorphism

I am trying to show that given $\phi: R\rightarrow R'$ is a ring homomorphism $R, R'$ are rings, there exist a homomorphism between the group of units that is $\phi': R^\times\rightarrow R'^\times$ is a group homomorphism. I was able to show that the the ring homomorphism maps units to units but I am not sure how to check the homomorphism condition explicitly.

• I am sorry.. what is your question? – user87543 Nov 17 '14 at 14:39
• The homomorphism condition for $\phi'$ is that $\phi'(xy) = \phi'(x)\phi'(y)$ for all units $x,y \in R^\times$, correct? You seem to be using the map $\phi$ restricted to $R^\times$ for your $\phi'$, so what do we know about the multiplicative properties of $\phi$? – Joshua Mundinger Nov 17 '14 at 14:41
• @Alqatrkapa: We know that $\phi(xy)=\phi(x)\phi(y)$ so this condition will also hold for our restricted map? – Rutherford Mark Nov 17 '14 at 14:45
• @RutherfordMark Precisely. – Joshua Mundinger Nov 17 '14 at 14:46

(I assume you're using $\phi'=\phi\rvert_{R^\times}$, that is, the restriction of $\phi$ to $R^\times$.)
Then, the statement follows trivially from the fact that $\phi$ is a ring homomorphism: From $$\forall x,y\in R.\quad \phi(x\cdot y)=\phi(x)\cdot\phi(y)$$ and $R^\times\subseteq R$, it is obvious that $$\forall x,y\in R^\times.\quad \phi'(x\cdot y)=\phi'(x)\cdot\phi'(y) \text.$$


text:
# Acceleration From Different Frames

If the acceleration of a particle, as seen from two frames $$S_1$$ and $$S_2$$, is of equal magnitude i.e. $$4 \ \text{ms}^{-2}$$ in both the cases, then which option is correct?

×


text:
Yeah.

By putting $\int_{0}^{\infty }{y^ne^{-xy}\,dx}$ which its result it's pretty much you showed. We can solve this via Gamma function.

This is not the only one paramater to solve integrals by constructing double ones, there're lots of tricks.


text:
# Circle A has a radius of 2  and a center of (2 ,7 ). Circle B has a radius of 6  and a center of (7 ,5 ). If circle B is translated by <-1 ,1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Feb 7, 2018

circle B do not overlap.
Shortest distance is $3.764 - 2 = 1.764$

#### Explanation:

Circle A
Center $\left(2 , 7\right)$
radius $2$

Equation of circle A
${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = {2}^{2}$

Circle B
Center $\left(7 , 5\right)$
radius $6$

Equation of circle B
${\left(x - 7\right)}^{2} + {\left(y - 5\right)}^{2} = {6}^{2}$

ranslation of B
<-1,1>
Center of B after translation $\left(\begin{matrix}7 - 1 \\ 5 + 1\end{matrix}\right)$
Center of B after translation $\left(6 , 6\right)$

Equation of circle B after translation
${\left(x - 6\right)}^{2} + {\left(y - 6\right)}^{2} = {6}^{2}$

Intersection of the circles
${\left(x - 2\right)}^{2} + {\left(y - 7\right)}^{2} = {2}^{2}$
and
${\left(x - 6\right)}^{2} + {\left(y - 6\right)}^{2} = {6}^{2}$
can be found out

Distance from center of Circle A to
center of Circle B is
$\left(2 , 7\right) - - - - \left(6 , 6\right)$
=sqrt((6-7)^2+(6-2)^2
=sqrt((1+4)
$= \sqrt{5}$
$= 2.236$

Radius of circle A is $2$
point on the Perimeter of the circle B is away from its center by 6
point on the Perimeter of the circle B is away from (2,7) is$6 - 2.236$
$= 3.764$
Hence, the two circles donot overlap.
Shortest distance is $3.764 - 2 = 1.764$

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text:
# Changes for page XWiki Enterprise 2.3 Milestone 1

<
From version
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on 2010/03/24
To version
edited by Caleb James DeLisle
on 2010/03/25
>
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## Details

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 ... ... @@ -33,6 +33,8 @@ 33 33 image:AnnotationsTab.png 34 34 ))) 35 35 36 +You can find a more detailed description of the annotations feature on the [[documentation page>>platform:Features.Annotations]]. 37 + 36 36 == Color Theme Editor == 37 37 38 38 The new editor allows fast and easy modifications for the look and feel of the wiki. You'll find in this editor ... ... @@ -55,6 +55,18 @@ 55 55 ))) 56 56 * The possibility to undo the latest modifications or reset all changes 57 57 60 +Read more about Color Themes on the dedicated [[application page>>code:Applications.ColorThemeApplication]]. 61 + 62 +== CAPTCHA support for anonymous comments == 63 + 64 +{{warning}}TODO{{/warning}} 65 + 66 +== Translations == 67 + 68 +* Newly supported language: Vietnamese 69 +* WYSIWYG newly supported language: Traditional Chinese 70 +* All the translations have been updated 71 + 58 58 = Known issues = 59 59 60 60 * [[Bugs we know about>>http://jira.xwiki.org/jira/secure/IssueNavigator.jspa?reset=true&&type=1&pid=10010&resolution=-1&sorter/field=updated&sorter/order=DESC]]
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Integers that Represent Different Situations

## Write variable expressions to represent word problems including positives and negatives

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Integers that Represent Different Situations
Credit: Edna Winti
Source: https://farm9.staticflickr.com/8581/15888043608_a954e00951.jpg

Aidin’s homework is to write about the world's highest geological feature. He discovered that Mount Everest reached 29,035 ft above sea level. Mauna Kea, only 13,796 ft above sea level. But since it started beneath the water, its total length from base to height was 33,476 ft. And the Mariana Trench is the deepest trench in the world at 36,070 ft! Which feature should Aidin write about?

In this concept, you will learn how to use integers to represent different situations in the real world.

### Integers to Represent Different Situations

There are many different kinds of numbers.

A whole number is any counting number. Numbers, like 1/2 or .75, that you use to describe a part of something are not whole numbers.

A positive number is a number greater than zero. It can be written with or without a + symbol in front of it. A gain in something is written with a positive number. Profit, increase, and income are all examples of words that mean positive.

A negative number is a number that is less than zero. It is always written with a - symbol in front of it. A loss is written with a negative number. Decrease, spend, and decline are examples of words that mean negative.

An integer is any positive whole number or its negative. Zero is also considered an integer.

Let's look at an example.

.

On this thermometer the temperature reads 100º. The temperature is above zero. That means that it is positive. Therefore, we can write it either as +100 or just as 100. A temperature that is below zero is negative. A temperature that is 5 degrees below zero, therefore, is written -5.

Here is another example.

Determine whether the integer given is positive or negative and write it out:

Sasha goes to the mall and spends $45.00 from her bank account at the clothing store. First, determine whether Sasha gains or loses money. Sasha "spends" this money, so she no longer has it. This makes it a loss. Then, write it as a negative integer. The answer is -45. ### Examples #### Example 1 Earlier, you were given a problem about Aidin and his science report. He has to choose between Mount Everest, at 29,035 ft above sea level, Mauna Kea which is 33,476 ft from its base to peak, but only 13,796 ft above sea level, and the Mariana Trench, which is actually the deepest point in the world, starting at 36,070 ft below sea level. His report is on the "highest" geological feature - which one does he write about? The "highest" geological feature is the one that stretches the farthest above sea level. In other words, the biggest integer. First, he writes them each as an integer. Mount Everest is 29,035 ft above sea level. "Above sea level" means it is positive. Then he writes it as a positive integer. Mount Everest is +29,035 Mauna Kea is trickier. It is 33,476 ft tall. But it is only 13,796 ft above sea level. That means it is 13,796ft positive. The rest of its length is beneath the water, below sea level, so it doesn't count. Then, he writes the part that is positive as a positive integer. Mauna Kea is +13,796 The Mariana Trench is 36,070 ft below sea level. "Below sea level" means that it is negative. Then, he writes it as a negative integer. The Mariana Trench is -36,070. Next, he writes them in a list to compare them. Mount Everest: +29,035 Mauna Kea: +13,796 The Mariana Trench: -36,070 Finally, he picks the one that is the biggest positive integer. That is Mount Everest. The answer is Mount Everest is the highest geological feature in the world. #### Example 2 People invest money in the stock market all the time. Sometimes the overall market increases and their money grows. Sometimes it decreases and they lose money. The amount of increase or decrease is characterized by positive or negative "points." Which of the following situations is expressed by a negative integer? 1. No gain or loss in the stock market There is no loss or gain. Therefore, the integer used to describe it is zero, which is neither negative nor is it positive. 1. The average plummeted 15 points The word "plummeted" indicates an extreme loss. Therefore, it would be written as -15. This is a negative integer, so it is the answer. 1. The overall rise was 30% The market is characterized as rising, which is positive. However, a percentage is not a whole number, so it is not an integer. Write out each of the following situations as an integer. #### Example 3 The head of the volcano is located 50 feet below sea level. First, determine whether the location is positive or negative. Sea level is zero in this example. That means that "below sea level" is negative. Then, write it as a negative integer. The answer is . #### Example 4 Rahul spent$100.00 on baseball cards.

First, determine whether Rahul gained or lost money.

He "spent" it. That means he no longer has it. So it is a loss.

Then, write it as a negative integer.

#### Example 5

Alaska experienced a heat wave of  on July 4th.

First, determine whether the temperature is above or below zero.

A "heat wave" means it's very hot. If it were below zero, it would be very cold. That means it is positive.

Then, write it as a positive integer.

### Review

Write an integer to represent each situation.

1. A loss of 20 points
2. A gain of 14 points
3. A profit of $20.00 4. A loss of$18.00
5. An elevation of 500 ft.
6. 200 feet below sea level
7. 8 degrees below zero
8. 78 degrees
9. A decrease of $68.00 10. An increase of$55.00
11. The dow did not rise or fall today.
12. The plane descended at a rate of 15%.
13. The submarine surfaced.
14. The cost and the earnings evened out.
15. The stock market had an overall rate of 18% above.

To see the Review answers, open this PDF file and look for section 11.1.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

Difference

The result of a subtraction operation is called a difference.

Integer

The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...

Negative Numbers

Negative numbers are numbers that are less than zero.

Positive Numbers

Positive numbers are numbers that are greater than zero.


text:
[1]:

import emat
emat.versions()

emat 0.5.2, plotly 4.14.3

# PRIM¶

Patient rule induction method (PRIM) is a scenario discovery algorithm that operates on an existing set of data with model inputs and outputs (i.e., you have already designed and run a set of experiments using either a core model or a meta-model.) Generally, a decently size set of experiments (hundreds or thousands) is used to describe the solution space, although no minimum number of experiments is formally required.

PRIM is used for locating areas of an outcome space that are of particular interest, which it does by reducing the data size incrementally by small amounts in an iterative process as follows:

• Candidate boxes are generated. These boxes represent incrementally smaller sets of the data. Each box removes a portion of the data based on the levels of a single input variable.
• For categorical input variables, there is one box generated for each category with each box removing one category from the data set.
• For integer and continuous variables, two boxes are generated – one box that removes a portion of data representing the smallest set of values for that input variable and another box that removes a portion of data representing the largest set of values for that input. The step size for these variables is controlled by the analyst.
• For each candidate box, the relative improvement in the number of outcomes of interest inside the box is calculated and the candidate box with the highest improvement is selected.
• The data in the selected candidate box replaces the starting data and the process is repeated.

The process ends based on a stopping criteria. For more details on the algorithm, see Friedman and Fisher (1999) or Kwakkel and Jaxa-Rozen (2016).

The PRIM algorithm is particularly useful for scenario discovery, which broadly is the process of identifying particular scenarios of interest in a large and deeply uncertain dataset. In the context of exploratory modeling, scenario discovery is often used to obtain a better understanding of areas of the uncertainty space where a policy or collection of policies performs poorly because it is often used in tandem with robust search methods for identifying policies that perform well (Kwakkel and Jaxa-Rozen (2016)).

## The Mechanics of using PRIM¶

In order to use PRIM for scenario discovery, the analyst must first conduct a set of experiments. This includes having both the inputs and outputs of the experiments (i.e., you’ve already run the model or meta-model).

[2]:

import emat.examples
designed = model.design_experiments(n_samples=5000, sampler='mc', random_seed=42)
results = model.run_experiments(designed, db=False)

In order to use PRIM for scenario discovery, the analyst must also identify what constitutes a case that is “of interest”. This is essentially generating a True/False label for every case, using some combination of values of the output performance measures as well as (possibly) the values of the inputs. Some examples of possible definitions of “of interest” might include:

• Cases where total predicted VMT (a performance measure) is below some threshold.
• Cases where transit farebox revenue (a performance measure) is above some threshold.
• Cases where transit farebox revenue (a performance measure) is above above 50% of budgeted transit operating cost (a policy lever).
• Cases where the average speed of tolled lanes (a performance measure) is less than free-flow speed but greater than 85% of free-flow speed (i.e., bounded both from above and from below).
• Cases that meet all of the above criteria simultaneously.

The salient features of a definition for “of interest” is that (a) it can be calculated for each case if given the set of inputs and outputs, and (b) that the result is a True or False value.

For this example, we will define “of interest” as cases from the Road Test example that have positive net benefits.

[3]:

of_interest = results['net_benefits']>0

Having defined the cases of interest, to use PRIM we pass the explanatory data (i.e., the inputs) and the ‘of_interest’ variable to the Prim object, and then invoke the find_box method.

[4]:

from emat.analysis.prim import Prim

[5]:

discovery = Prim(
of_interest,
threshold=0.02,
scope=scope
)

[6]:

box1 = discovery.find_box()

The find_box command actually provides a number of different possible boxes along a (heuristically) optimal trajectory, trading off coverage against density. We can plot a static figure showing the tradeoff curve using the show_tradeoff command:

[7]:

ax = box1.show_tradeoff()

This figure shows the trade-off between coverage and density.

• Coverage is percentage of the cases of interest that are in the box (i.e., number of cases of interest in the box divided by total number of cases of interest). The starting point of the PRIM algorithm is the unrestricted full set of cases, which includes all outcomes of interest, and therefore, the coverage starts at 1.0 and drops as the algorithm progresses.
• Density is the share of cases in the box that are case of interest (i.e., number of cases of interest in the box divided by the total number of cases in the box). As the box is reduced, the density will increase (as that is the objective of the PRIM algorithm).

For the statistically minded, this tradeoff can also be interpreted as the tradeoff between Type I (false positive) and Type II (false negative) error. High coverage minimizes the false negatives, while high density minimizes false positives.

By default, the PRIM algorithm sets the “selected” box position as the particular box at the end of the peeling trajectory (highest density, at the top of the top of the tradeoff curve), which has the highest density, but the smallest coverage.

[8]:

box1

[8]:

<PrimBox peel 82 of 82>
coverage: 0.21257
density:  0.99649
mean: 0.99649
mass: 0.05700
restricted dims: 6
min        max
expand_capacity      0.004812  60.324206
input_flow                125        150
value_of_time        0.067766   0.238765
beta                 4.505424    5.49961
amortization_period        18         50
alpha                 0.10917   0.198906

In a Jupyter notebook, the analyst can interactively view and select other possible boxes using the tradeoff_selector method. This method opens a dynamic viewer which shows the same basic figure as the static one above, but when hovering over each possible tradeoff point, a pop-up box appears with more details about that point, including which dimensions are restricted, and what the restrictions are. A selection of the preferred tradeoff can be made by clicking on any point. The currently selected preferred tradeoff is marked with an “X” instead of a dot.

[9]:

box1.tradeoff_selector()

In addition to the point-and-click interface, the analyst can also manually set the selection programmatically using Python code, selecting a preferred box (by index) that trades off density and coverage.

[10]:

box1.select(30)
box1

[10]:

<PrimBox peel 31 of 82>
coverage: 0.72904
density:  0.71043
mean: 0.71043
mass: 0.27420
restricted dims: 3
min       max
expand_capacity  0.004812  86.06928
input_flow            116       150
value_of_time    0.067171  0.238765

To help visualize these restricted dimensions better, we can generate a plot of the resulting box, overlaid on a ‘pairs’ scatter plot matrix (splom) of the various restricted dimensions.

In the figure below, each of the three restricted dimensions represents both a row and a column of figures. Each of the off-diagonal charts show bi-dimensional distribution of the data across two of the actively restricted dimensions. These charts are overlaid with a green rectangle denoting the selected box. The on-diagonal charts show the relative distribution of cases that are and are not of interest (unconditional on the selected box).

[11]:

box1.splom()

Depending on the number of experiments in the data and the number and distribution of the cases of interest, it may be clearer to view these figures as a heat map matrix (hmm) intead of a splom.

[12]:

box1.hmm()

## Non-Rectangular Boxes¶

Regardless of any analyst intervention in box selection using the select method, initial box identified using PRIM will always be a rectangle (or, more generally, a hyper-rectangle). Every restricted dimension is specified independently, not conditional on any other dimension.

A second box can be overlaid after the first to expand the solution to include a non-rectangular area. The second tier box itself is still another rectangle, but it can be defined using different boundaries, possibly on different dimensions, to give in aggregate a non-convex shape. Creating such a second box is done by calling the find_box method again, after the particulars of the first box are finalized.

[13]:

box2 = discovery.find_box()

[14]:

box2

[14]:

<PrimBox peel 48 of 48>
coverage: 0.11751
density:  0.43014
mean: 0.43014
mass: 0.07300
restricted dims: 8
min                max
unit_cost_expansion          95.000837         142.588934
expand_capacity               0.004812          97.595439
input_flow                         112                150
beta                          4.051416            5.49961
value_of_time                 0.035604           0.238765
amortization_period                 20                 50
alpha                         0.110845           0.199972
debt_type            {Rev Bond, Paygo}  {Rev Bond, Paygo}

The cases considered for the second box are only those cases not within the first box. Because our first box had a relatively high coverage and high density, we have already captured most of the cases of interest and not too many other cases, so the second box has a much diminished trade-off curve. The maximum value of the coverage for this second box is one minus the selected final coverage of the first box.

[15]:

box2.show_tradeoff();

We can also make a similar scatter plot for this second box. There are 8 restricted dimensions in the currently selected solution, which can result in an unwieldy large scatter plot matrix. To help, the splom command allows you to override the selection of dimensions to display in the rows and/or columns of the resulting figure, by explicitly giving a set of dimensions to display. These don’t necessarily need to be restricted dimensions, they can be any dimensions available in the original PRIM analysis.

Because this is the second box, the method will only plot the points that remain (i.e. the ones not included in the first box). This results in a “blotchy” appearance for may of the graphs in this figure, as most of the “of interest” cases were captured previously.

[16]:

box2.splom(
cols=['beta','input_flow','value_of_time','debt_type','yield_curve']
)

## PRIM API¶

class emat.analysis.Prim(x, y, threshold=0.05, *args, scope=None, explorer=None, **kwargs)[source]

Bases: emat.workbench.analysis.prim.Prim, emat.analysis.discovery.ScenarioDiscoveryMixin

Patient rule induction method

This implementation of Prim is derived from the EMA workbench, and is enhanced to work interactively within the TMIP-EMAT framework.

Parameters: x (pandas.DataFrame) – The independent variables, generally the scoped input parameters (uncertainties and/or policy levers) from a design of experiments. y (array-like) – The dependent variable, generally a pandas.Series or other one dimensional array with size equal to the number of rows in x. threshold (float) – The density threshold that a box must meet. obj_function ({LENIENT1, LENIENT2, ORIGINAL}) – The objective function used by PRIM. This defaults to a lenient objective function based on the gain of mean divided by the loss of mass. peel_alpha (float, default 0.05) – The parameter controlling the peeling stage. paste_alpha (float, default 0.05) – The parameter controlling the pasting stage. mass_min (float, default 0.05) – The minimum mass of a box. threshold_type ({ABOVE, BELOW}) – Whether to look above or below the threshold value. mode (RuleInductionType) – This PRIM implementation defaults to binary classification, but this argument can be used to switch to regression mode. In binary mode, the y array should only have binary values, while in regression mode it can contain other floating point values. update_function ({‘default’, ‘guivarch’}) – The default behavior for this algorithm after finding the first box is to remove all points in the first box, so that subsequent boxes do not include them. An alternate procedure suggested by Guivarch et al (2016) doi:10.1016/j.envsoft.2016.03.006 is to simply set all points in the first box to be no longer of interest. This alternate procedure is only valid in binary mode.
property boxes

Property for getting a list of box limits

boxes_to_dataframe(include_stats=False)

convert boxes to pandas dataframe

Parameters: include_stats (bool, default False) – If True, the box statistics will also be retrieved and returned in the same DataFrame as the boxes.
determine_coi(indices)[source]

Given a set of indices on y, how many cases of interest are there in this set.

Parameters: indices (ndarray) – a valid index for y the number of cases of interest. int ValueError – if threshold_type is not either ABOVE or BELOW
find_box()[source]

Execute one iteration of the PRIM algorithm.

This method will find one box, starting from the current state of Prim (i.e., over all observations that are not already within a previously selected PrimBox). All existing boxes will be “frozen” by this command, so that those prior boxes can no longer be modified by their select methods or from within their tradeoff_selector.

Returns: emat.analysis.PrimBox
show_boxes(together=False)

display boxes

Parameters: together (bool, otional) –
property stats

property for getting a list of dicts containing the statistics for each box

stats_to_dataframe()

convert stats to pandas dataframe

to_emat_boxes(scope=None)

Export boxes to emat.Boxes

Parameters: scope (Scope) – The scope Boxes
tradeoff_selector(n=- 1, colorscale='viridis', figure_class=None)[source]

Visualize the trade off between coverage and density.

This visualization plots all of the points along the peeling trajectory for the selected PrimBox, plotting coverage along the x axis, density along the y axis, and showing the number of restricted dimensions by color.

Coverage is percentage of the cases of interest that are in the box (i.e., number of cases of interest in the box divided by total number of cases of interest). The starting point of the PRIM algorithm is the unrestricted full set of cases, which includes all outcomes of interest, and therefore, the coverage starts at 1.0 and drops as the algorithm progresses.

Density is the share of cases in the box that are case of interest (i.e., number of cases of interest in the box divided by the total number of cases in the box). As the box is reduced, the density will increase (as that is the objective of the PRIM algorithm).

Parameters: n (int, optional) – The index number of the PrimBox to use. If not given, the last found box is used. If no boxes have been found yet, an initial box is found using the find_box method, and in this case giving any value other than -1 will raise an error. colorscale (str, default ‘viridis’) – A valid color scale name, as compatible with the color_palette method in seaborn. plotly.FigureWidget
class emat.analysis.PrimBox(prim, box_lims, indices)[source]

Bases: emat.workbench.analysis.prim.PrimBox

Information for a specific Prim box.

By default, the currently selected box is the last box on the peeling trajectory, unless this is changed via PrimBox.select().

box_number()[source]

The number of PRIM iterations up to and including this PrimBox.

Returns: int
drop_restriction(uncertainty='', i=- 1)[source]

Drop the restriction on the specified dimension for box i

Parameters: i (int, optional) – defaults to the currently selected box, which defaults to the latest box on the trajectory uncertainty (str) –

Replace the limits in box i with a new box where for the specified uncertainty the limits of the initial box are being used. The resulting box is added to the peeling trajectory.

explore(scope=None, data=None)[source]

Create or access the existing visualizer.

Parameters: scope (Scope, optional) – Override the exploratory scope for this Box. data (DataFrame, optional) – Override the data for this Box. emat.Visualizer
hmm(rows=None, cols=None)[source]

Generate a heatmap matrix showing this CartBox.

Parameters: rows (list-like, optional) – The dimensions to display as rows and columns of the resulting heatmap matrix. If not provided, each defaults to the set of restricted dimensions on the current CartBox. cols (list-like, optional) – The dimensions to display as rows and columns of the resulting heatmap matrix. If not provided, each defaults to the set of restricted dimensions on the current CartBox. plotly.FigureWidget
inspect(i=None, style='table', **kwargs)[source]

Write the stats and box limits of the user specified box to standard out. if i is not provided, the last box will be printed

Parameters: i (int, optional) – the index of the box, defaults to currently selected box style ({‘table’, ‘graph’}) – the style of the visualization kwargs are passed to the helper function that (additional) – the table or graph (generates) –
resample(i=None, iterations=10, p=0.5)[source]

Calculate resample statistics for candidate box i

Parameters: i (int, optional) – iterations (int, optional) – p (float, optional) – DataFrame
select(i)[source]

Select an entry from the peeling and pasting trajectory.

This will update the PRIM box to this selected box, as well as update the tradeoff_selector and explorer, if either is attached to this PrimBox.

Parameters: i (int) – The index of the box to select.
show_pairs_scatter(i=None, dims=None, markers=None)[source]

Make a pair wise scatter plot of all the restricted dimensions with color denoting whether a given point is of interest or not and the boxlims superimposed on top.

Parameters: i (int, optional) – dims (list of str, optional) – dimensions to show, defaults to all restricted dimensions seaborn PairGrid
show_ppt()[source]

show the peeling and pasting trajectory in a figure

show_tradeoff(cmap=<matplotlib.colors.ListedColormap object>)[source]

Visualize the trade off between coverage and density. Color is used to denote the number of restricted dimensions.

Parameters: cmap (valid matplotlib colormap) – a Figure instance
splom(rows=None, cols=None)[source]

Generate a scatter plot matrix showing this PrimBox.

Parameters: rows (list-like, optional) – The dimensions to display as rows and columns of the resulting scatter plot matrix. If not provided, each defaults to the set of restricted dimensions on the current CartBox. cols (list-like, optional) – The dimensions to display as rows and columns of the resulting scatter plot matrix. If not provided, each defaults to the set of restricted dimensions on the current CartBox. plotly.FigureWidget
to_emat_box(i=None, name=None, src_name=None)[source]

Convert the current PrimBox to a standard emat.Box.

Returns: emat.Box
to_json()[source]

Write out the current PrimBox in json format.

Returns: str
tradeoff_selector(colorscale='viridis', figure_class=None)[source]

Visualize the trade off between coverage and density.

This visualization plots all of the points along the peeling trajectory for this PrimBox, plotting coverage along the x axis, density along the y axis, and showing the number of restricted dimensions by color.

Coverage is percentage of the cases of interest that are in the box (i.e., number of cases of interest in the box divided by total number of cases of interest). The starting point of the PRIM algorithm is the unrestricted full set of cases, which includes all outcomes of interest, and therefore, the coverage starts at 1.0 and drops as the algorithm progresses.

Density is the share of cases in the box that are case of interest (i.e., number of cases of interest in the box divided by the total number of cases in the box). As the box is reduced, the density will increase (as that is the objective of the PRIM algorithm).

Parameters: colorscale (str, default ‘viridis’) – A valid color scale name, as compatible with the color_palette method in seaborn. FigureWidget
update(box_lims, indices)[source]

update the box to the provided box limits.

Parameters: box_lims (DataFrame) – the new box_lims indices (ndarray) – the indices of y that are inside the box
write_ppt_to_stdout()[source]

write the peeling and pasting trajectory to stdout


text:
## aNY Group Title plzzzz help me just one question A=72 ft2 B=74 ft2 C=86 ft2 D=32 ft2 Find the area of the figure. one year ago one year ago

• This Question is Open
1. aNY Group Title

2. whpalmer4 Group Title

Decompose that into a rectangle and a right triangle by extending the 6 ft line down to the base. Find the area of both of those figures and add them.

3. aNY Group Title

HMM

4. whpalmer4 Group Title

Can you find the base and the height of that triangle? The area will be 1/2 * base * height.

5. aNY Group Title

still don'y get it.

6. aNY Group Title

you there

7. aNY Group Title

anyone that can help

8. zepdrix Group Title

|dw:1359845754154:dw|Understand how we're starting this? Breaking it up into 2 easier shapes :)

9. aNY Group Title

yes

10. zepdrix Group Title

|dw:1359845881864:dw|

11. zepdrix Group Title

|dw:1359845947895:dw|

12. zepdrix Group Title

See how the dimensions of the triangle come about? :)

13. aNY Group Title

yaa

14. zepdrix Group Title

|dw:1359846112405:dw|So we want to find the area of these two objects separately. Then we'll add those area's together to get the total area of the figure.

15. aNY Group Title

ok so i got C

16. zepdrix Group Title

Hmm did you forget the formula for area of a triangle maybe? :o

17. zepdrix Group Title

cuz C is too big :c

18. aNY Group Title

idk what the answer is then

19. aNY Group Title

lol :)

20. zepdrix Group Title

well it feels like you're just guessing :) lol

21. aNY Group Title

i multiplide 10 and 7 i got 70 then i multiplide 4 and 2 then i got 16 then i add them together which i got 86

22. zepdrix Group Title

The 70 looks good. $$4 \times 2 \ne 16$$ , might wanna visit Mr Calculator for that one! :O And then remember, The area of a triangle we take half of the base times height.

23. aNY Group Title

woops i ment 8

24. zepdrix Group Title

Cool c: Just remember for a triangle,$\large A=\frac{1}{2}b\cdot h$

25. zepdrix Group Title

So we only want half of that 8 c:

26. aNY Group Title

hmm

27. zepdrix Group Title

|dw:1359846808816:dw|If we had a square like this, the area would be 8, because 4 x 2 = 8.

28. zepdrix Group Title

|dw:1359846855993:dw|But we have a triangle, which is half of that area.

29. zepdrix Group Title

still confused? <:o

30. aNY Group Title

yes

31. zepdrix Group Title

aw :c

32. aNY Group Title

ya sorry i'm not that smart

33. aNY Group Title

was 74

34. zepdrix Group Title

im sure that's not true c: math isn't for everyone!

35. zepdrix Group Title

ya 74 D:

36. aNY Group Title

so is 74 right

37. aNY Group Title

@zepdrix

38. zepdrix Group Title

yes c:


text:
A Breakdown in Simultaneity

Three enemy spacecraft have been causing trouble in the asteroid belt. They always travel in a line, evenly spaced apart, attempting to chase down local spacecraft to steal their goods. The local asteroid colonists have decided to set a trap to capture these three spacecraft. They'll get them to chase one of their fastest ships into an asteroid with a large hole in it and, once the three enemy ships are inside, close two giant trapdoors on each side of the asteroid to catch them. These spacecraft all travel close to the speed of light so the locals will have to take relativity into account. Intelligence about the enemy spacecraft reveals that, in their reference frame, they always travel 90 m behind their teammate, each spacecraft is 10 m in length, and their maximum velocity is 90% the speed of light (relative to the asteroids). The asteroid tunnel is only 215 m in length. In this problem we will analyze whether the locals will be able to capture the enemy spacecraft after taking into account relativity.

If the spacecraft are traveling at 90% the speed of light, what is the total length of the three-spacecraft team as observed from the asteroid?

My attempt was putting the number in the L=Lo( 1/$$\sqrt{}1-v2/c2$$

Then i add them up... but wasnt right... anyone has idea?!!!?! please!!

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