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

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

8

step-by-step explanation: