Using the given set of numbers find the mean (rounded to the nearest tenth), median,mode and range: 18,19,16,18,14,17,18 and 15
2 answers:
You might be interested in
Answer:
She will get <u>80mg</u> of dextromethorphan and <u>800mg</u> of guaifenesin. And the bottle last for <u>6 days</u> approximately.
Step-by-step explanation:
Given that the Robitussin DM contains dextromethorphan 10mg/5mL and gualfenesin 100mg/5mL. And we are also given that Mrs Smith took four doses and each dose is 2 teaspoons=2X5=10mL.
So, four doses=4X10=40mL.
So, dextromethorphan in 4 doses is =
And Guaifenesin in 4 doses is =
Dosage of medicine daily she has to take=40mL and the bottle contains 237 mL. Hence the number of days bottle last = ≈6 days approximately.
Answer:
There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.
Step-by-step explanation:
For each passenger, there are only two outcomes possible. Either they show up for the flight, or they do not show up. This means that we can solve this problem using binomial distribution probability concepts.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinatios of x objects from a set of n elements, given by the following formula.
And is the probability of X happening.
A probability is said to be unusually low if it is lower than 5%.
For this problem, we have that:
There are 200 reservations, so .
A passenger consists in a passenger not showing up. There is a .0995 probability that a passenger with a reservation will not show up for the flight. So .
Find the probability that when 200 reservations are accepted for United Flight 15, there are more passengers showing up than there are seats available.
X is the number of passengers that do not show up. It needs to be at least 18 for the flight not being overbooked. So we want to find , with
. We can use a binomial probability calculator, and we find that:
.
There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.