75 percent of the 60 passengers disembarked at the terminal.

a. how many passengers disembarked at the terminal.

b. what percent of the passengers did not disembark at the terminal?

Okay Well

A.

75% Of 60 = 45

B.  

Since 25% + 75% = 100%

Then 25% Of 60 = 15

Hope That Helps!

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;