Assume each newborn baby has a probability of approximately 0.55 of being female and 0.45 of being male. for a family with three children, let x equal number of children who are girls. a. identify the three conditions that must be satisfied for x to have the binomial distribution. b. identify n and p for the binomial distribution. c. find the probability that the family has one girl and two boys.

1. There must be exactly two possible outcomes (girl, boy).

2. Repeated trials must be independent.

3. The probabilities of the two outcomes don't change from trial to trial.

n = 3 (there are 3 children "generated")

p = 0.55 (the probability that a child is a girl)

The probability that a family of 3 consists of one girl and two boys is

In general, the probability that a binomially distributed random variable X is equal to r is

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

50- 250 is correct i did that one too

step-by-step explanation: