2 5/9. Because you first find how many times 9 goes into 23 and then subtract the sum of it from the denominator and keep the same denominator as your answer.
Answer:
what are you asking? Because I probably have the answer
Answer: 99.51%
Step-by-step explanation:
Given : A survey found that women's heights are normally distributed.
Population mean : 
Standard deviation: 
Minimum height = 4ft. 9 in.=
Maximum height = 6ft. 2 in.=
Let x be the random variable that represent the women's height.
z-score : 
For x=57, we have

For x=74, we have

Now, by using the standard normal distribution table, we have
The probability of women meeting the height requirement :-

Hence, the percentage of women meeting the height requirement = 99.51%
Since x can be any value as long as the denominator equals 0 (it doesn't matter if it's positive or negative), we have to figure out when x^2=0, which is when x=0. Therefore, the domain is (-inf, 0) U (0, inf)