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# linear-transformation-eye-catch

• Linear Transformation $T(X)=AX-XA$ and Determinant of Matrix Representation Let $V$ be the vector space of all $n\times n$ real matrices. Let us fix a matrix $A\in V$. Define a map $T: V\to V$ by $T(X)=AX-XA$ for each $X\in V$. (a) Prove that $T:V\to V$ is a linear transformation. (b) Let $B$ be a basis of $V$. Let $P$ be the matrix […]
• Symmetric Matrix and Its Eigenvalues, Eigenspaces, and Eigenspaces Let $A$ be a $4\times 4$ real symmetric matrix. Suppose that $\mathbf{v}_1=\begin{bmatrix} -1 \\ 2 \\ 0 \\ -1 \end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $1$ of $A$. Suppose that the eigenspace for the eigenvalue $2$ is $3$-dimensional. (a) Find an […]
• Hyperplane in $n$-Dimensional Space Through Origin is a Subspace A hyperplane in $n$-dimensional vector space $\R^n$ is defined to be the set of vectors $\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\in \R^n$ satisfying the linear equation of the form $a_1x_1+a_2x_2+\cdots+a_nx_n=b,$ […]
• Questions About the Trace of a Matrix Let $A=(a_{i j})$ and $B=(b_{i j})$ be $n\times n$ real matrices for some $n \in \N$. Then answer the following questions about the trace of a matrix. (a) Express $\tr(AB^{\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\trans}$ is the transpose matrix of […]
• Prove that the Length $\|A^n\mathbf{v}\|$ is As Small As We Like. Consider the matrix $A=\begin{bmatrix} 3/2 & 2\\ -1& -3/2 \end{bmatrix} \in M_{2\times 2}(\R).$ (a) Find the eigenvalues and corresponding eigenvectors of $A$. (b) Show that for $\mathbf{v}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}\in \R^2$, we can choose […]
• Find the Rank of the Matrix $A+I$ if Eigenvalues of $A$ are $1, 2, 3, 4, 5$ Let $A$ be an $n$ by $n$ matrix with entries in complex numbers $\C$. Its only eigenvalues are $1,2,3,4,5$, possibly with multiplicities. What is the rank of the matrix $A+I_n$, where $I_n$ is the identity $n$ by $n$ matrix. (UCB-University of California, Berkeley, […]
• Common Eigenvector of Two Matrices and Determinant of Commutator Let $A$ and $B$ be $n\times n$ matrices. Suppose that these matrices have a common eigenvector $\mathbf{x}$. Show that $\det(AB-BA)=0$. Steps. Write down eigenequations of $A$ and $B$ with the eigenvector $\mathbf{x}$. Show that AB-BA is singular. A matrix is […]


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A total of $60,000 was invested for one year. Part of this [#permalink] ### Show Tags 05 Aug 2006, 07:09 3 This post was BOOKMARKED 00:00 Difficulty: 65% (hard) Question Stats: 58% (02:40) correct 42% (01:37) wrong based on 162 sessions ### HideShow timer Statistics A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x ?

(1) $$x = \frac{3y}{4}$$
(2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.
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26 Aug 2010, 10:26
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udaymathapati wrote:
A total of $60,000 was invested for one year. But of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the$60,000 for that year was
$4,080, what is the value of x? (1) x = (3/4) y (2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2. Let the amount invested at x% be $$a$$, then the amount invested at y% would be $$60,000-a$$. Given: $$a\frac{x}{100}+(60,000-a)\frac{y}{100}=4,080$$ (we have 3 unknowns $$x$$, $$y$$, and $$a$$). Question: $$x=?$$ (1) $$x=\frac{3}{4}y$$ --> $$y=\frac{4x}{3}$$ --> $$a\frac{x}{100}+(60,000-a)\frac{4x}{3*100}=4,080$$ --> still 2 unknowns - $$x$$ and $$a$$. Not sufficient. (2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2 --> $$\frac{a}{60,000-a}=\frac{3}{2}$$ --> $$a=36,000$$ --> $$36,000*\frac{x}{100}+(60,000-36,000)\frac{y}{100}=4,080$$ --> still 2 unknowns - $$x$$ and $$y$$. Not sufficient. (1)+(2) From (1) $$a\frac{x}{100}+(60,000-a)\frac{4x}{3*100}=4,080$$ and from (2) $$a=36,000$$ --> only 1 unknown - $$x$$, hence we can solve for it. Sufficient. Answer: C. _________________ Senior Manager Joined: 14 Jul 2005 Posts: 402 Followers: 1 Kudos [?]: 27 [0], given: 0 ### Show Tags 05 Aug 2006, 09:28 D From 1 => x/100 * 60,000 + 4x/3 * 100 * 60,000 = 4080 From 2 => 3a + 2a = 4080 => a = 816 Hence x/100 * 60,000 = 3 * 816 Hence you can solve for x using either equation Manager Joined: 20 Mar 2006 Posts: 200 Followers: 1 Kudos [?]: 3 [0], given: 0 Re: investment [#permalink] ### Show Tags 05 Aug 2006, 10:31 ong wrote: A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?
(1)x = 3y/4
(2)The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.

(1) gives relationship between x and y. we still dont know the number of months to get a numerical value for x
Insuff BCE
(2) gives the ratio of Interest from x/Interest from y. Cannot determine x from this. CE
(1) & (2) No new information about the number of months

E

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05 Aug 2006, 10:35
[quote="gmatornot"]D

From 1 => x/100 * 60,000 + 4x/3 * 100 * 60,000 = 4080
(1) I thought SI formula was prinicpal * rate * number of months/100. How are you getting x/100 * 60000.
(2) From the question stem i understand that part of the 60000 was invested @ x% and the rest @ y% ?
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05 Aug 2006, 14:40
heman wrote:
gmatornot wrote:
D

From 1 => x/100 * 60,000 + 4x/3 * 100 * 60,000 = 4080
(1) I thought SI formula was prinicpal * rate * number of months/100. How are you getting x/100 * 60000.
(2) From the question stem i understand that part of the 60000 was invested @ x% and the rest @ y% ?

You are RIGHT. I completely misread the question. I believe the answer is E as you have identified...
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05 Aug 2006, 15:30
ong wrote:
A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the$60,000 for that year was $4,080, what is the value of x? (1)x = 3y/4 (2)The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2. Let part of amount = A, then the other part = 60000-A A*(x/100) + (60000-A)*(y/100) = 4080 S1. x=3y/4 A*(3y/400) + (60000-A)*(y/100) = 4080 We need value A to solve this. => insufficient. S2. A*(x/100) = (3/5)*4080 = 2448 (60000-A)*(y/100) = (2/5)*4080= 1632 We need value A to solve this. => insufficient. Combine S1 and S2. A*(3y/400) = 2448 => A=2448*400/3y=326400/y (60000-A)*(y/100) = 1632 We have two variables with two eq. thus sufficient. Hence, C.. (600-3264/y)*(y) = 1632 600y - 3264 = 1632 600y = 1632+3264 600y = 4896 y = 8.16 % x=3y/4 = 8.16*3/4 = 6.12% _________________ The only thing that matters is what you believe. Senior Manager Joined: 22 May 2006 Posts: 371 Location: Rancho Palos Verdes Followers: 1 Kudos [?]: 114 [0], given: 0 Re: investment [#permalink] ### Show Tags 05 Aug 2006, 15:33 heman wrote: (1) & (2) No new information about the number of months E Heman If the total interest earned by the$60,000 for that year was $4,080, what is the value of x _________________ The only thing that matters is what you believe. Manager Joined: 20 Mar 2006 Posts: 200 Followers: 1 Kudos [?]: 3 [0], given: 0 Re: investment [#permalink] ### Show Tags 05 Aug 2006, 15:48 freetheking wrote: heman wrote: (1) & (2) No new information about the number of months E Heman If the total interest earned by the$60,000 for that year was $4,080, what is the value of x I still dont get it. Wouldnt it be possible for amount A @ x% for n months + amount (60000 - A) @ y % for (12-n) months = 4080. Then won't we have different value for x and y depending on the time period of investment within a year? Or does the ? mean that each part invested @ x% and y% was for a year? I had a similar solution to you except I had a variable for time period and so 2 eqns and 3 variables. Heman Senior Manager Joined: 22 May 2006 Posts: 371 Location: Rancho Palos Verdes Followers: 1 Kudos [?]: 114 [0], given: 0 Re: investment [#permalink] ### Show Tags 05 Aug 2006, 15:52 ong wrote: A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?

x and y are simple annual interest..
dealing with one year investment.. not monthly..
heman wrote:
Or does the ? mean that each part invested @ x% and y% was for a year?
Yes!
Heman

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19 Jan 2008, 09:01
marcodonzelli wrote:
A total of $60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the$60,000 for that year was $4,080, what is the value of x? (1) x = 3/4 y (2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2. I think it is C First statement is not suff because we don't know which amount was exactly invested, second is not suff since we don't know the rates but combined we can solve it II. 3/5*60000=36000 I. 36000*3/4y+24000*y=4,080 17/4y=4080/12000 => y=8% => x=8*3/4 =6% Manager Joined: 10 Aug 2009 Posts: 130 Followers: 3 Kudos [?]: 67 [0], given: 10 Re: DS: Interests ratios [#permalink] ### Show Tags 23 Aug 2009, 19:54 1)n1+ n2 = 60000 n1- amount invested at x%, n2 - amount invested at y%. 2) n1*x%+n2*y% = 4080 To solve this for x, we need two things: 1) the ratio of n1 to n2 (given in the second statement) 2) the ration of x to y (given in the first statement) So, both statements together are sufficient. Senior Manager Joined: 30 Aug 2009 Posts: 286 Location: India Concentration: General Management Followers: 3 Kudos [?]: 164 [0], given: 5 Re: annual interest [#permalink] ### Show Tags 29 Dec 2009, 07:31 kirankp wrote: A total of$60,000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned by the $60,000 for that year was$4,080, what is the value of x?

(1) x = 3y/4

(2) The ratio of the amount that earned interest at the rate of x percent per year to the amount that earned interest at the rate of y percent per year was 3 to 2.

will go with C

let "a" be the amount which gets Simple interest at rate of x%/yr. Then SI = ax/100
then "60000 -a " is the amount which gets Simple interest at rate of y%/yr. Then SI = (60000-a)y/100

so we have (ax)/100 + (60000-a)y/100 = 4080

from stmnt 1- we have x = 3y/4 . even after substituting value of x in above equation we will not be able to get the value of x. Insuff

from stmnt 2 - we have (ax/100)/ (60000-a)y/100 =3/2. Again not suff

taking together x = 3y/4 and (ax/100)/ (60000-a)y/100 =3/2 we will get the value of x. hence suff
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13 May 2011, 12:40
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13 May 2011, 18:50
a - x% interest

(60000 - a) - y% interest

ax/100 + (60000 - a)y/100 = 4080

(1)

x = 3y/4

Not Sufficient

(2)

a/(60000 - a) = 3/2

No information about x and y, insufficient.

(1) + (2)

Sufficient

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13 May 2011, 19:37
b tells nothing about relative information between x and y.

a+b p can be calculated first.
together with x and y information, x can be found out.

C
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13 May 2011, 19:38
$$px/100 + (60-p)y/100 = 4080$$

1. Not sufficient

we dont know anything about amount invested at x%

2. Not sufficient.

not enough to find p or x or y.

Together ,

its sufficient

as we can find p,y and x

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Re: A total of $60,000 was invested for one year. Part of this [#permalink] ### Show Tags 24 Jul 2014, 15:59 Let A be the part that gets interest at the rate of x% B = 60000 -A that gets interest at y%. There are two unknowns in question stem - the relation between A and B and the relation between x and y Statement 1 : gives relation between x and y, but we do not know relation between A and B --> insufficient Statement 2 gives A:B = 3:2, --> A = 36000 and B = 24000 but nothing about x:y, --> Insufficient Combining 1 and 2 --> Can be solved as equation can be reduced to one variable , hence sufficient Re: A total of$60,000 was invested for one year. Part of this   [#permalink] 24 Jul 2014, 15:59
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# Why are we allowed to make algebraic calculation on units

Is there math proof that we can cancel out units in Physics? For example:

$$\require{cancel}distance = \frac{meters}{\cancel{second}} * \cancel{second}$$.

So we see that seconds cancel out and we left with meters which is correct but how is it possible if it is not actually division (meters per second) it is only our interpretation of speed. Why are we allowed to make algebraic calculations on units? Also, we can use dimensional analysis to check that we do everything correctly but why? Why does it always work?

• See en.wikipedia.org/wiki/Buckingham_%CF%80_theorem and references therein.
– PPR
Nov 11 '20 at 13:07
• "it is not actually division" -- who says it isn't? Nov 11 '20 at 15:04
• If you want full-on mathematical rigor, I can't really explain it than Terence Tao did. Nov 11 '20 at 15:19
• I found this from four years ago: math.stackexchange.com/questions/1980010/… (Why do units from physics behave like numbers?) A lot to digest there. But for your question about dimensional analysis, check out @Jack M's pendulum example. Basically, the result e.g. for period of a pendulum should not depend on the units used to measure the quantities it depends on. Nov 11 '20 at 15:53
• My recommendation is to always use dimensional analysis to check answers. If your answer is dimensionally consistent, it has a chance of being correct; if on the other hand your answer is dimensionally inconsistent, it is definitely wrong. Nov 11 '20 at 16:43

Good question. I don't have any elegant, deep mathematical answer for you. But let's consider an object moving at 3 m/sec and consider a time interval of 5 sec. We can lay out the object's progress along a line consisting of 5 sections, in each of which the object has moved 3 m. So the total distance moved is 3m + 3m +3m +3m +3m = 15m. Knowing multiplication, we can also find the total distance as 3m x 5 = 15m. Notice that in this last calculation time doesn't play a role. I'm simply multiplying a distance by a pure number. So (3 m/sec) x (5 sec) is really 3m x 5. The seconds cancel. That to me would justify treating the sec unit like a number to be canceled.

• Thank you for your answer. it makes sense for this specific example but I'm not sure that it will work for others especially for more difficult ones Nov 13 '20 at 16:09

The question seems trivial, but it isn't at all!

Some strange formal properties of algebraic calculations on units (e.g. the elusive rad unit which appears or disappears) should suggest that "something is wrong" instead of the more usual W.Allen's "Whatever works".

The problem concerns the logic (syntax and semantics) of the ''mathematical'' language of physical quantities. What does mean "the product of a force times a distance" or "the division of a distance by a time"? And what is the rigorous logic of the dimensional algebra?

Units of measure are samples of quantities, but the calculations made with symbols of units are a particular form of dimensional calculus. This (little-shared) statement is not the result of an improvisation: here I am forced to be self-referential by quoting the document:

http://pangloss.ilbello.com/Fisica/Metrologia/grf.pdf

Unfortunately the text is not in English language. I emphasize S.3.7 ''Algebra of units of measure'' and the last sentences:

The usefulness of dimensional monomials consists in making the calculation of the relationships between the various units of measurement of physical quantities algebrically intuitive, conforming it to the ordinary rules of algebra (as illustrated in the examples).
The ordinary calculations with units of measure are based on dimensional relations between classes of quantities, they are not univocal relations between units of measure. We must not deceive ourselves to be able to obtain in this way the exact conventional name or even the definition of the calculated unit!

As Pangloss points out, this is surprisingly non-trivial!

The sort of manipulations you are talking about are part of a quantity calculus. This is not to be confused with Newton's and Leibniz's Calculus. A calculus is just a method for computation. This is a method of manipulating quantities which yield "the same" quantity with different units.

Your method of "canceling" the units is the normal way of phrasing one of the rules that can be applied in quantity calculus. And, we show very informally that it works, because nobody has found a way it doesn't work. It's anthropocentric, but its the truth. Like so many systems, we can do the things we can do in that system because we evolved countless systems that didn't work, and this one did.

A major aspect of answering your "why" question is the axiomization of unit conversions. This is the process of "turning the crank" on these unit manipuations to produce all true unit conversions. In theory, if you could do this, you could start to make statements about how they work and prove something more profound.

As it turns out, we cant do that. We do not currently believe that units "in general" can be axiomized in this way. However, we are confident that the basic scientific units (meters, seconds, etc.) can be treated this way. Perhaps they are a special case, perhaps they are a general rule.

Metrologia keeps their papers behind a paywall, but a particularly interesting paper might be On quantity calculus and units of measurement. This discusses an interesting pattern where our basic scientific units can be treated as quantities in their own right (i.e. "ft" corresponds to "1 ft"), and thus admit algebraic operations like division. However, it points out that, in general, this is not a valid construction.

The rules of algebra apply to the manipulation of symbols. The expressions for the units of measure are symbols.

• Only a subset of typical algebraic rules applies to units or physical values, respectively. I cannot arbitrarily add them for instance. Also, how do you define a mathematical symbol other than something to which the rules of algebra apply? And why are units such symbols and not something else? Nov 11 '20 at 15:24


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Best April deals - hours only!Up to 80% off on all courses and bundles.-Close
Introduction
Simple CASE WHEN
6. Simple CASE WHEN with NULLs
Searched CASE WHEN
CASE WHEN with aggregates
CASE WHEN with GROUP BY
Summary

## Instruction

That's correct! Having gone through Part 5 about NULLs, you might want to ask: okay, but how do NULLs behave in CASE WHEN?

The answer is pretty straightforward: NULL will always go to the ELSE condition if you include it. Let's check this out in an exercise.

## Exercise

For table candidate, show each candidate's id, preferred_contact and the third column which will print:

• 'traditional' if the preferred_contact is 'mail'.
• 'modern' if it's 'mobile' or 'e-mail'.
• 'other' otherwise.

Look what is shown for NULL values.


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# A pipeline tapers from 500 mm diameter to 250 mm diameter. From this pipe, water is flowing at a volume of 6.4 m3/s. Find the average water velocity at the small end.

This question was previously asked in
UPPCL JE CE Offical Paper I 2014
View all UPPCL JE Papers >
1. 212.44 m/s
2. 100.44 m/s
3. 157.44 m/s
4. 130.44 m/s

Option 4 : 130.44 m/s
Free
CT 1: Network Theory 1
11207
10 Questions 10 Marks 10 Mins

## Detailed Solution

Given data,

Discharge in pipe Q = 6.4 m3/sec

From the Equation of continuity,

As we know,

A.V. = Q = constant

Discharge in the pipe at any point is constant

Hence,

Velocity at smaller end = $$\frac{Discharge}{Area of small end}$$

$$\frac{6.4}{\frac{A}{4}(0.25)^2}$$

= 130.45 m/sec


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# Boehmians

Not to be confused with Bohemians.

In mathematics, Boehmians are objects obtained by an abstract algebraic construction of "quotients of sequences." The original construction was motivated by regular operators introduced by T. K. Boehme. Regular operators are a subclass of Mikusiński operators, that are defined as equivalence classes of convolution quotients of functions on ${\displaystyle [0,\infty )}$. The original construction of Boehmians gives us a space of generalized functions that includes all regular operators and has the algebraic character of convolution quotients. On the other hand, it includes all distributions eliminating the restriction of regular operators to ${\displaystyle [0,\infty )}$.

Since the Boehmians were introduced in 1981, the framework of Boehmians has been used to define a variety of spaces of generalized functions on ${\displaystyle \mathbb {R} ^{N}}$ and generalized integral transforms on those spaces. It was also applied to function spaces on other domains, like locally compact groups and manifolds.

## The general construction of Boehmians

Let ${\displaystyle X}$ be an arbitrary nonempty set and let ${\displaystyle G}$ be a commutative semigroup acting on ${\displaystyle X}$. Let ${\displaystyle \Delta }$ be a collection of sequences of elements of ${\displaystyle G}$ such that the following two conditions are satisfied:

(1) If ${\displaystyle (\phi _{n}),(\psi _{n})\in \Delta }$, then ${\displaystyle (\phi _{n}\psi _{n})\in \Delta }$,

(2) If ${\displaystyle x,y\in X}$ and ${\displaystyle \phi _{n}x=\phi _{n}y}$ for some ${\displaystyle (\phi _{n})\in \Delta }$ and all ${\displaystyle n\in \mathbb {N} }$, then ${\displaystyle x=y}$.

Now we define a set of pairs of sequences:

${\displaystyle {\mathcal {A}}=\{((x_{n}),(\phi _{n})):x_{n}\in X,(\phi _{n})\in \Delta ,\phi _{m}x_{n}=\phi _{n}x_{m}{\text{ for all }}m,n\in \mathbb {N} \}}$.

In ${\displaystyle {\mathcal {A}}}$ we introduce an equivalence relation:

${\displaystyle ((x_{n}),(\phi _{n}))}$ ~ ${\displaystyle ((y_{n}),(\psi _{n}))}$ if ${\displaystyle \phi _{m}y_{n}=\psi _{n}x_{m}{\text{ for all }}m,n\in \mathbb {N} }$.

The space of Boehmians ${\displaystyle {\mathcal {B}}(X,\Delta )}$ is the space of equivalence classes of ${\displaystyle {\mathcal {A}}}$, that is ${\displaystyle {\mathcal {B}}(X,\Delta )={\mathcal {A}}/}$~.

## References

• J. Mikusiński, Operational Calculus, Pergamon Press (1959).
• T. K. Boehme, The support of Mikusiński operators, Trans. Amer. Math. Soc. 176 (1973), 319–334.
• J. Mikusiński and P. Mikusiński, Quotients de suites et leurs applications dans l'analyse fonctionnelle (French), [Quotients of sequences and their applications in functional analysis], C. R. Acad. Sci. Paris Sr. I Math. 293 (1981), 463-464.
• P. Mikusiński, Convergence of Boehmians, Japan. J. Math. (N.S.) 9 (1983), 159–179.


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# A lift whose cage is 3 meter high is moving up with an acceleration of 2 meter per second square . A piece of stone is dropped from the top of the cage of the lift when it's velocity is 8 meter per second . If gram = 10 meter per second second square , then the stone will reach the floor of the lift after

dhruba
17 Points
one year ago
listing the facts …
set the point from where the ball was thrown as origin and follow sign convention
initial velocity of the ball(u)= +8 m/s(not -8m/s)
height to be covered(s)=-3m
acceleration is only due to gravity(g)_=-10 m/s^2

using equation $x=ut+1/2at^2$  we get  -3=8(t)+1/2 (-10)(t^2)   solving u get t=3/5 sec or 1 sec

dhruba
17 Points
one year ago
saying position would be more accurate wherever i told height ….........thats why we could give the sign negativ eor positive so please do the corrcetion


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#### What do the graphs represent?

The first two graphs show the number of temporal reproduction (denoted $$\mathcal{R}(t)$$), i.e., at a given time $$t$$, the average number of people an infected person infects over the course of his or her infection. It is therefore an estimate of the potential spread of the epidemic. The shaded areas indicate the confidence interval and the line is the median.

The third graph represents the incidence data, i.e. here the number of new cases detected each day. Depending on the choice in the menu, this incidence can refer to screenings, hospitalizations, ICU admissions, or deaths.

The fourth graph allows you to zoom in on a particular time period by dragging the small sliders (move the left one to the right to see the recent values better).

#### If $$\mathcal{R}(t)$$ is smaller than one, then everything’s okay?

Not necessarily: $$\mathcal{R}(t)$$ represents the trend of the epidemic but it does not reflect its current state. For instance, an epidemic can be declining but there can hundreds of thousands of infected people and overwhelmed ICU services.

#### Why two different charts for $$\mathcal{R}(t)$$?

These two graphs are based on two slightly different methods, which are described in the Model tab.

The EpiEstim method is less sensitive to time variations (it calculates a 3-day average) and uses a more detailed model to forecast recent values.

The R0 method is more sensitive to changes in the incidence curve and the most recent value can have a significant effect on the curve.

#### Is one of the two methods to be preferred?

Yes, but it depends on the piece of information you are interested in.

The R0 method is to be preferred if you are moving away from the present — and therefore if you are looking for an estimate for a past date.

Conversely, for a recent estimate of $$\mathcal{R}(t)$$ we will rather turn to the method EpiEstim.

#### Which serial interval to choose?

The serial interval is used to estimate the average number of days between the date of symptoms onset in an “infecter” and the date of symptoms onset in an “infectee”. This data is needed to measure the reproduction number. Unfortunately, the serial interval is still largely unknown for epidemics in France and Europe.

We therefore offer the user to choose between several serial interval distributions measured on COVID-19 epidemics in Asia or estimated in the light of previous epidemics.

#### Why do the incidence data differ slightly from those found on other websites?

In order to overcome the weekend reporting period for new cases, the data were smoothed using a 7-day moving average.

Each point is therefore the mean of the last 7 days.

For purists, we used the function \texttt{zoo::rollmean(x, k = 7, align = "right", fill = NA)}.

Note that this smoothing produces decimal values in the incidence series.

#### Why aren’t there $$\mathcal{R}(t)$$ estimates for more recent dates?

There is always a lag between the state of the epidemic and what can be estimated from the data. This gap is due to the fact that the events observed (screening, hospitalisation, entry into intensive care units, death) only occurs a certain number of days after individuals have been infected — the event most likely to characterise the epidemic in real time, but in practice never known.

These time lags vary from one individual to another, but for the sake of simplicity we have set them at times that we consider to be representative based on a model from our team (Sofonea et al., medRxiv):

• 10 days lag for newly detected cases;
• 12 days lag for hospitalizations;
• 14 days lag for the ICU admissions;
• 28 days lag for deaths.

#### What is the most reliable type of data?

Incidence data on PCR detection are generally the least reliable because they are very sensitive to variations in detection policies. For example, in France sampling was limited at the beginning of the epidemic but now lots of tests are being performed. This increase in screening mechanically translates into an increase in the number of cases detected, and therefore an increase in the reproduction number, even though the epidemic could be strongly decreasing.

The most reliable data for measuring $$\mathcal{R}(t)$$ is therefore the one where the screening policy has varied the least over time. In France, this would therefore correspond to data on new ICU admissions. However this data is not always available.

#### What is the difference between departments and countries?

National data aggregates departmental data. This addition makes them less sensitive to stochastic fluctuations. The estimates are therefore more robust with narrower confidence intervals.

On the other hand, if only a few departments account for the majority of cases, then the national data may poorly reflects the situation of the epidemic in the least affected departments.

#### Why are not all countries/regions/departments in the proposed lists?

In order to be processed by the packages used, the data needs to satisfy several conditions:

• the first data point is non-zero;
• there are no missing values;
• the number of consecutive observations must be greater than eight.