There are four blood types, and not all are equally likely to be in blood banks. In a certain blood bank, 49% of donations are type O blood, 27% of donations are type A blood, 20% of donations are type B blood, and 4% of donations are type AB blood. A blood bank wants to know how many donations will be required, on average, to achieve at least one of each of the blood types. Is it appropriate to use the geometric distribution to calculate probabilities in this situation?

Yes, the geometric distribution is appropriate.
No, since each trial is not independent of the other trials.
No, because it is not looking for the first occurrence of success.
No, since a success and failure on each trial cannot be defined.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

{3 + i, 3 - i, 3, -1}

step-by-step explanation:

12cm

the answer is d