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
Her scores would be 98, 82, and 81. Their sum would be 261 then you would divide it be 3 and it would give you 87 then to find the median line the numbers up from greatest to least and 82 would be in the middle. To find the range subtract 81 from 98 and it gives you 17. Did this help?
The answer is 4.9. 32/26=6/x, cross multiple to get 32x=156, divide by 32 to get x=4.875~4.9
They are pairs on the corridinant
To solve this problem we would plug 58 into the equation for a
Making the equation 59÷58
59÷58=1.0172413793103
Or
1.02
