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

hope this .

r3t40

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)