Assume the blood type distribution to be a: 41 %, b: 9 % , ab: 4%, o: 46 % . what is the probability that the blood of randomly selected individual will contain either the a antigen or the b antigen? that it will contain neither the a nor the b antigen? 2.

 a) 50% , and b) 50%.

Step-by-step explanation:

Since we have given that

P(A) = 41%

P(B) = 9%

P(AB) = 4%

P(O) = 46%

Since antigen A or antigen B are independent events.

So, P(A ∪ B) = P(A) + P(B)

Probability neither the A nor the B is given by

Hence, a) 50% , and b) 50%.

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1