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
Volume of cube=side³
ok, so you need to know the difference or sum of cubes
a³+b³=(a+b)(x²-xy+y²)
so
(4p)³+(2q²)³=
(4p+2q²)((4p)²-(4p)(2q²)+(2q²)²)=
(4p+2q²)(16p²-8pq²+4q⁴)
3rd option
Answer: 20 unit.
Step-by-step explanation:
Since, Here the vertices of the rhombus defg are d(1, 4), e(4, 0), f(1, –4), and g(–2, 0).
Where, de, ef, fg, gd are sides of the rhombus defg.
By the distance formula,





Thus, the side of rhombus = 5
By the property of rhombus,
de = ef = fg = gd = 5 unit.
Thus, the perimeter of the given rhombus defg = de + ef + fg + gd = 5+5+5+5 = 20 unit
890×(1+0.187÷12)^(12)−890×(1+0.125÷12)^(12)
=63.61 saved