Find the probability of drawing 1 snicker first then the probability of drawing another snicker after that, then multiply them together. For the first choice, the probability of drawing a Snickers is 28/33 (snickers/total candy bars). For the second choice, there are only 32 candy bars left, 27 of which are Snickers. So that probability is 27/32, so...
The probability of BOTH things happening is their product 756/1,056 which is 63:88
There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area.
only A and E satisfy both inequality, in the overlapping shaded area.
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true.
for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
-4≥-4&1/3 this is valid as well. So Yes, A satisfies both inequalities.
Do the same for point E (5,-4)
Part C: the line y<-2x+4 is a dotted line going through (0,4) and (-2,0)
the shaded area is below the line
farms A, B, and D are in this shaded area.
A.
The graph of y= -1/2+5 is incorrect.
