At a certain hospital, the probability of a child being born female is 0.51. If 80 babies are born on the a certain day, the probability that more than half of these babies are female is closest to? a. 0.61, b. 0.70, c. 0.39, d. 0.47, 0.53 (I will award you the brainiest)

Answer :

0.53

Step-by-step explanation:

Given the following :

Probability of female = probability of success on a single trial = 0.51

Number of babies = 80

Probability that more than half of the babies are female : p(X >80/2) = P(X > 40)

The problem can be solved using the binomial probability function :

P(X > 40) = [ p(X= 41) + p(X= 42) + +p(X= 80)]

In other to save computation time, we can use the binomial probability calculator

Hence ; P(X > 40) = 0.527 = 0.53

yes   because as the functions are graphed they eventually will intersect unless otherwise

step-by-step explanation:

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb