There are two 5's in this number.
The first 5, is in the ones place, which would have the value of 5
The second 5, is in the hundredths place, which would have the value of 5/100, simplified 1/20
Hope this helps :)
education Roman Catholic School one five two weeks
Answer:
okay..hahah will do..
edit- wait how old are, are you a little kid, I can't follow you if you are a little kid...
Answer:
the probability that the graduate program will have enough funding for all student that join the program is 0.3653 (36.53%)
Step-by-step explanation:
since each student is independent on others the random variable X= x students of 45 applicants will join the program has a binomial probability distribution
P(X=x)= n!/[(n-x)!*x!]*p^x*(1-p)^x
where
n= total number of students= 45
p= probability that a student join the program= 0.7
x= number of students that join the program
then in order to have enough funding x should not surpass 30 students , then
P(X≤30)= ∑P(X) for x from 1 to 30 = F(30)
where F(30) is the cumulative probability distribution
then from binomial probability tables
P(X≤30)= F(30)= 0.3653 (36.53%)
therefore the probability that the graduate program will have enough funding for all student that join the program is 0.3653 (36.53%)
Answer:
The point estimate for the proportion is p = 0.4725
The 95% confidence interval for the proportion of non-fatal accidents that involved the use of a cell phone is (0.4236, 0.5214).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:
![n = 400](https://tex.z-dn.net/?f=n%20%3D%20400)
Point estimate
![\pi = p = \frac{189}{400} = 0.4725](https://tex.z-dn.net/?f=%5Cpi%20%3D%20p%20%3D%20%5Cfrac%7B189%7D%7B400%7D%20%3D%200.4725)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4725 - 1.96\sqrt{\frac{0.4725*0.5275}{400}} = 0.4236](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.4725%20-%201.96%5Csqrt%7B%5Cfrac%7B0.4725%2A0.5275%7D%7B400%7D%7D%20%3D%200.4236)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4725 + 1.96\sqrt{\frac{0.4725*0.5275}{400}} = 0.5214](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.4725%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.4725%2A0.5275%7D%7B400%7D%7D%20%3D%200.5214)
The 95% confidence interval for the proportion of non-fatal accidents that involved the use of a cell phone is (0.4236, 0.5214).