Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701
You can use a square plus b square equal C square to fijd each length . or you can memorize the special triangles or you can also do proportional problem. it all works
Answer:
Part a) <1=72°
Part b) <2=108°
Part c) <3=72°
Part d) <4=108°
Step-by-step explanation:
step 1
Find the measure of angle 1
we know that
<1+108°=180° -----> by supplementary angles
so
<1=180°-108°=72°
step 2
Find the measure angle 2
we know that
<2=108° -----> by corresponding angles
step 3
Find the measure angle 3
we know that
<3=<1-----> by corresponding angles
so
<3=72°
step 4
Find the measure angle 4
we know that
<4=108° -----> by alternate exterior angles
Answer:
the answer is c
Step-by-step explanation:
