X^2-7x+38=5x+3
x^2-7x+38-5x-3=5x+3-5x-3
x^2-12x+35=0
Factoring:
(x-5)(x-7)=0
Two solutions:
x-5=0→x-5+5=0+5→x=5
x-7=0→x-7+7=0+7→x=7
Answer: x=5 and x=7
Answer: Options B. 7 and D. 5
A) 5 to be chosen among a Total : 10 Men + 8 Women
¹⁸C₅ = (18!)/(5!)(13!) = 8,568 groups of five
b) A must to have men and women. If so we have to deduct all groups of 5 that are all men and all group of 5 that are all women
Groups of 5 with only men: ¹⁰C₅ = 252
Groups of 5 with only women: ⁸C₅ = 56
So number of committees of 5 men and women mixed =
8568 - 252 - 56 = 8,260 committees
c) 3 Women and 2 Men:
⁸C₃ x ¹⁰C₂ = 2,520 groups of 3 W and 2 M
d) More women than men, it means:
3 W + 2 M OR (we have found it in c) = 2,520)
4 W + 1 M OR ⁸C₄ x ¹⁰C₁ →→→→ = 700
5 W + 0 M OR ⁸C₅ x ¹⁰C₀ →→→→ = 56
Total where W>M = 3,276 groups of 5 where women are at least 3
Answer:
Step-by-step explanation:
here is a solution :
I believe the answer is 2