(a) there are 8C2 = 28 ways of picking 2 girls from 8
And there are 21C4 = 5985 ways of picking 4 boys
Required number of ways for 2g / 4b = 28 * 5985 = 167,580
(b) at least 2 girls means combinations of 2g/4b , 3g,3b , 4g/2b , 5g 1b or
6 girls.
2g/4b = 167,580 ways
3g/3b = 8C3 * 21C3 = 56 * 1330 = 74,480
4g/2b = 8C4* 21C2 = 70 * 210 = 14,700
5g 1b = 8C5* 21 = 56*21 = 1176
6 girls = 8C6 = 28
adding these up we get the answer to (b) which is 257,964
Answer:
x<-11/26
Step-by-step explanation:
1/5(2/3x-1/2)>x+1/3
2/15x-1/10>x+1/3
2/15x-x>1/3+1/10
-13/15x>11/30
x<(11/30)(-15/13)
x<-11/26
Answer:
Fraction Form:
Decimal Form: 0.4583
Step-by-step explanation:
Apply the fraction rule:
Least Common Multiple (LCM) of 12 and 8: 24
Apply the fraction rule:
Answer:
a possible solution is (10,1)
Step-by-step explanation:
Here, we want to solve the given inequality and determine one possible solution
From the question, we have it that;
x + y ≤ 13
Mathematically, we have it that 1 dime is worth $0.1
A nickel is worth $0.05
0.1x + 0.05y ≥ 0.95
please check the attachment for the graph of both equations
The point of intersection represents the solution.
we have a point here as (10,1)