First question seems incomplete :
Answer:
40 ways
Step-by-step explanation:
Question B:
Number of boys = 6
Number of girls = 4
Number of people in committee = 3
Number of ways of selecting committee with atleast 2 girls :
We either have :
(2 girls 1 boy) or (3girls 0 boy)
(4C2 * 6C1) + (4C3 * 6C0)
nCr = n! ÷ (n-r)!r!
4C2 = 4! ÷ 2!2! = 6
6C1 = 6! ÷ 5!1! = 6
4C3 = 4! ÷ 1!3! = 4
6C0 = 6! ÷ 6!0! = 1
(6 * 6) + (4 * 1)
36 + 4
= 40 ways
The answer I get
8=24
No solution
Answer:
19/25 = 0.76
311/500 = 0.622
5/8 = 0.625
145/8 = 18.12500
Step-by-step explanation:
Divide the first number by the second like 145 divided by 18.
-3-24=-27, -3x-24=72
Answer is -3 and -24
Let's eliminate x. To achieve this, mult. the 2nd equation by 2, obtaining
-8x + 2y = 10. Now add this result to the first equation:
8x-2y= -10
-8x + 2y = 10
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0
The two equations actually represent the same line. Without an x- or a y-value, we cannot solve for the value of the other variable. There are infinitely many solutions.
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