From 2f558eee0ac9cf08944308015d6f9c5ea8195131 Mon Sep 17 00:00:00 2001
From: 8388c1425d3a684d1ec014af187ba020
 <8388c1425d3a684d1ec014af187ba020@app-learninglab.inria.fr>
Date: Tue, 24 Mar 2020 12:08:54 +0000
Subject: [PATCH] Update toy_document_orgmode_python_en.org

---
 module2/exo1/toy_document_orgmode_python_en.org | 6 ++----
 1 file changed, 2 insertions(+), 4 deletions(-)

diff --git a/module2/exo1/toy_document_orgmode_python_en.org b/module2/exo1/toy_document_orgmode_python_en.org
index 058e29b..a7082ea 100644
--- a/module2/exo1/toy_document_orgmode_python_en.org
+++ b/module2/exo1/toy_document_orgmode_python_en.org
@@ -24,8 +24,6 @@ from math import *
 pi
 #+end_src
 
-#+RESULTS:
-
 * 2. *Buffon's needle
 Applying the method of [[https://en.wikipedia.org/wiki/Buffon%27s_needle_problem][_Buffon's needle_]], we get the *approximation*
 #+begin_src python :results output :exports both
@@ -37,7 +35,7 @@ theta=np.random.uniform(size=N, low=0, high = pi/2)
 2/(sum((x+np.sin(theta))>1)/N)
 #+end_src
 
-#+RESULTS:
+
 
 * 3.  Using a surface fraction argument
 A method that is easier to understand and does not make use of the sin
@@ -65,7 +63,7 @@ plt.savefig(matplot_lib_filename)
 print(matplot_lib_filename)
 #+end_src
 
-#+RESULTS:
+
 
 It is then straightforward to obtain a (not really good) approximation
 of \pi by counting how many time, on average, $X^2$ + $^{}Y^2$ is smaller
-- 
2.18.1