Since the problem states "at least" we need to also find probability of 3 H or 4 H or 5 H
Now find the probability of flipping a head 4 times;

⁴
= (1/16)
Now probability of flipping a head 3 times: (4C3)(1/2)⁴ = 4/16
Probability of flipping a head 2 times; (4C2)(1/2)⁴=6/16
(1/16)+(4/16)+(6/16)=11/16
Probability of flipping a fair coin 4 times with at least 2 heads is 11/16.
Hope I helped :)
Hello,
f(x)=2x^3-x^2-4x+9=(((0+2)x-1)x-4)x+9
f(3)=(((0+2)*3-1)*3-4)*3+9=((2*3-1)*3-4)-3+9=(5*3-4)*3+9=11*3+9=33+9=42
or
f(3)=2*3^3-3^2-4*3+9=2*27-9-12+9=54-12=42
I believe that there are 55 1/3 cup servings with an extra 5/12.
It doesn’t show the bar graph