The math club at a certain school has 10 members, of which 6 are seniors and 4 juniors. in how many ways can they form a group of 5 members to go to a tournament, if at least 4 of them have to be seniors (aka either a group of 4 seniors and 1 junior, or a group of 5 seniors

66 ways

Step-by-step explanation:

Given;

Number of senior math club members = 6

Number of junior math club members = 4

Total number of members of the club = 10

To form a group of 5 members with at least 4 seniors.

N = Na + Nb

Na = number of possible ways of selecting 4 seniors and 1 junior

Nb = number of possible ways of selecting 5 seniors.

Since the selection is does not involve ranks(order is not important)

Na = 6C4 × 4C1 = 6!/4!2! × 4!/3!1! = 15 ×4 = 60

Nb = 6C5 = 6!/5!1! = 6

N = Na + Nb = 60+6

N = 66 ways

statement #3: proof- ang is congruent to ang ecd : angles are congruent.  

step-by-step explanation:

1: the missing angles in statement #3  

a.) acb  

reason- the reason why the answer is acb is because angle ecd and angle acb are both the same size *that's what the word congruent means this is really simple.  

2: the missing word in reason #3  

c.) vertical  

reason- the reason why the answer is vertical is because the angles in that diagram on figure 2 is slanted, which is vertical, and the e and a both lay on vertical lines. along with the other's, but e and a are the 2 main points you want to focus on. and c is kind of like your starting point, because c is for both angles, not just one.  

i hope that was and understanding.

really your just stupid if you don't know this

step-by-step explanation: