Answer: 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
-5x + 3y = 19
-5x + 7y = 11
-1(-5x + 3y = 19) ==> 5x - 3y = -19
1(-5x + 7y = 11) ==> -5x - 7y = 11
5x - 3y = -19
-5x - 7y = 11
-10y = -8
/-10 /-10
y = 4/5 = 0.8
-5x + 3y = 19
-5x + 3(0.8) = 19
5x + 2.4 = 19
- 2.4 - 2.4
5x + = 16.6
/5 /5
x = 3.32
-5x + 3y = 19
-5x + 3(4/5) = 19
5(5x + 3(4/5) = 19)
25x + 12 = 95
- 12 - 12
25x = 83
/25 /25
x = 83/25
- Hope that helps!
PS: the x answers are the same thing, one is in fraction, while the other one is in decimal.
Answer:
x - -5.78526086, -1.50769051, 2.29295138
Step-by-step explanation:
Try to graph each side of the equation. The solution Is probably the x-value of the point of intersection. Sorry if this is wrong I tried my best!
Step by step explanation:
Original Equation: 43 - 3m; m = 6
Substitute 6 in for m
43 - 3(6)
Multiply 3*6
43 - 18
Subtract 43 - 18
25
Your answer is: 25