Hey Marc-king, good to see you again. Here are the solutions to the exercise: 1) Calculer la moyenne des notes des élèves de cette classe à cette évaluation. To calculate the mean (average) grade, we sum the product of each grade and its corresponding number of students, then divide by the total number of students. Step 1: Calculate the total number of students. $$N = 7 + 4 + 6 + 2 + 4 = 23$$ Step 2: Calculate the sum of (grade $\times$ number of students). $$\sum (\text{note} \times \text{nombre d'élèves}) = (8 \times 7) + (9 \times 4) + (10 \times 6) + (12 \times 2) + (15 \times 4)$$ $$= 56 + 36 + 60 + 24 + 60$$ $$= 236$$ Step 3: Calculate the mean grade. $$\text{Moyenne} = \frac{\sum (\text{note} \times \text{nombre d'élèves})}{N} = \frac{236}{23} \approx 10.26$$ The mean grade is $\boxed{\frac{236}{23} \approx 10.26}$. 2.a) Recopier et compléter ce tableau par ligne des effectifs cumulés croissants et par celle des effectifs cumulés décroissants. Here is the completed table: | Note sur 20 | 8 | 9 | 10 | 12 | 15 | |---|---|---|---|---|---| | Nombre d'élèves | 7 | 4 | 6 | 2 | 4 | | Effectifs cumulés croissants (ECC) | 7 | 11 | 17 | 19 | 23 | | Effectifs cumulés décroissants (ECD) | 23 | 16 | 12 | 6 | 4 | Effectifs cumulés croissants (ECC): For note 8: 7 For note 9: $7 + 4 = 11$ For note 10: $11 + 6 = 17$ For note 12: $17 + 2 = 19$ For note 15: $19 + 4 = 23$ Effectifs cumulés décroissants (ECD): For note 15: 4 For note 12: $4 + 2 = 6$ For note 10: $6 + 6 = 12$ For note 9: $12 + 4 = 16$ For note 8: $16 + 7 = 23$ b) En déduire la note médiane. The total number of students is $N = 23$. Since $N$ is odd, the median is the value of the $\frac{N+1}{2}$-th observation. Step 1: Calculate the position of the median. $$\text{Position de la médiane} = \frac{23+1}{2} = \frac{24}{2} = 12$$ The median is the 12th grade when the grades are ordered. Step 2: Find the grade corresponding to the 12th position using the cumulative increasing frequencies. 7 students have a grade of 8 or less. 11 students have a grade of 9 or less. 17 students have a grade of 10 or less. The 12th student falls into the category of students who scored 10. The median grade is $\boxed{10}$. 3) On choisit au hasard et simultanément 5 élèves de la classe. a) Déterminer le nombre de choix possibles. This is a combination problem, as the order of selection does not matter. We are choosing 5 students from a total of 23 students. Step 1: Use the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. $$\binom{23}{5} = \frac{23!}{5!(23-5)!} = \frac{23!}{5!18!}$$ $$= \frac{23 \times 22 \times 21 \times 20 \times 19}{5 \times 4 \times 3 \times 2 \times 1}$$ $$= \frac{23 \times 22 \times 21 \times 20 \times 19}{120}$$ $$= 23 \times 11 \times 7 \times 1 \times 19$$ $$= 33649$$ The number of possible choices is $\boxed{33649}$. b) Déterminer le nombre de choix possibles contenant exactement trois élèves ayant au moins 10 à cette évaluation. First, we need to identify the number of students who scored at least 10 and the number of students who scored less than 10. Step 1: Count students with a grade of at least 10. Grades $\ge 10$: 10, 12, 15 Number of students: $6 + 2 + 4 = 12$ students. Step 2: Count students with a grade less than 10. Grades $< 10$: 8, 9 Number of students: $7 + 4 = 11$ students. (Total students: $12 + 11 = 23$, which is correct). Step 3: Calculate the number of ways to choose exactly 3 students from the 12 students with a grade of at least 10. $$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$$ Step 4: Calculate the number of ways to choose the remaining $5-3=2$ students from the 11 students with a grade less than 10. $$\binom{11}{2} = \frac{11!}{2!(11-2)!} = \frac{11 \times 10}{2 \times 1} = 11 \times 5 = 55$$ Step 5: Multiply the results from Step 3 and Step 4 to find the total number of choices. $$\text{Nombre de choix} = \binom{12}{3} \times \binom{11}{2} = 220 \times 55 = 12100$$ The number of possible choices is $\boxed{12100}$. Send me the next one 📸