A football team carried out a report to see the impact of stretching on preventing injury. Of the 32 footballers in the squad 25 stretch regularly. Of those who stretch, 3 got injured last year. There was a total of 8 injured players last year. The results can be presented in a frequency tree. What fraction of players who do stretch get injured?

853.3

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

step-by-step explanation:

-1   1/6.   you dont subtract the denominators

step-by-step explanation: