From 589d5cc5dc71a92b774fca99ee42d6c9e3d584ae Mon Sep 17 00:00:00 2001
From: 87599ea463bc80856e7cb0d6ce6fa3a1
 <87599ea463bc80856e7cb0d6ce6fa3a1@app-learninglab.inria.fr>
Date: Wed, 25 Oct 2023 22:24:30 +0000
Subject: [PATCH] Update toy_document_en.Rmd

---
 module2/exo1/toy_document_en.Rmd | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/module2/exo1/toy_document_en.Rmd b/module2/exo1/toy_document_en.Rmd
index c33e052..69eb6b4 100644
--- a/module2/exo1/toy_document_en.Rmd
+++ b/module2/exo1/toy_document_en.Rmd
@@ -11,14 +11,14 @@ knitr::opts_chunk$set(echo = TRUE)
 ```
 
 ## Asking the maths library
-My computer tells me that $\pi$ is approximatively
+My computer tells me that $\pi$ is *approximatively*
 
 ```{r}
 pi
 ```
 
-## Buffon’s needle
-Applying the method of [Buffon’s needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the approximation
+## Buffon's needle
+Applying the method of [Buffon's needle](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem), we get the **approximation**
 
 ```{r}
 set.seed(42)
@@ -30,7 +30,7 @@ theta = pi/2*runif(N)
 
 ## Using a surface fraction argument
 
-A method that is easier to understand and does not make use of the $\sin$ function is based on the fact that if $X\sim U(0,1)$  and $Y\sim U(0,1)$, then $P[X^2+Y^2\leq 1]=\pi /4$ (see [“Monte Carlo method” on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
+A method that is easier to understand and does not make use of the $\sin$ function is based on the fact that if $X\sim U(0,1)$  and $Y\sim U(0,1)$, then $P[X^2+Y^2\leq 1]=\pi /4$ (see ["Monte Carlo method" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method)). The following code uses this approach:
 
 ```{r}
 set.seed(42)
-- 
2.18.1