(3c² + 2d)(–5c² + d)-------->(3c²)*(-5c²)+(3c²)*(d)+(2d)*(-5c²)+(2d)*(d)
then
(3c²)*(-5c²)-------------> -15c^4
(3c²)*(d)----------------> +3dc²
(2d)*(-5c²)-------------> -10dc²
(2d)*(d)----------------> +2d²
[(3c² + 2d)(–5c² + d)]=[-15c^4+3dc²-10dc²+2d²]-------> [-15c^4-7dc²+2d²]
12xy
sqrt144=12 sqrtx^2=x sqrty^2=y
Answer:
The probability that all three have type B+ blood is 0.001728
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they have type B+ blood, or they do not. The probability of a person having type B+ blood is independent of any other person. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
The probability that a person in the United States has type B+ blood is 12%.
This means that 
Three unrelated people in the United States are selected at random.
This means that 
Find the probability that all three have type B+ blood.
This is P(X = 3).


The probability that all three have type B+ blood is 0.001728
Ones because it is the last place value