1. Let's call:
x: the width of the walkway.
b: 2x+16 (there is a widht "x" of the walkway on each side).
h: 2x+32 (there is a widht "x" of the walkway on each side).
A: 924 ft^2 ( the area of the pool of including the walkway).
2. The area of a rectangle is:
A=(b)(h)
b: the base of the rectangle (2x+16)
a: the height of the rectangle (2x+32)
3. Then, we have a quadratic equation:
924=(2x+16)(2x+32)
4x^2+64x+32x+512-924=0
4x^2+96x-412=0
4. We apply the quadratic formula to find the value of "x":
x=(-b±√(b^2-4ac))/2a
x=3.71
The answer is: The width of the walkway is 3.71 ft
Yes they can if 7 + 7 equals more then 11 the. The answer is yes
The complete factor from:
2x3 + 6x2 + 10x + 30
2x(x+3) + 10(x+3)
(x+3)(2x+10)
2(x+5)(x+3)
hope this help
I have an expression

floating around in my head; let's see if it makes sense.
The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

Plugging in the numbers,

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.
So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.