Answer:
The 99% confidence interval for the mean time required by all individuals to compete the form is between 17.168 minutes and 30.032 minutes.
Step-by-step explanation:
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 50 - 1 = 49
Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of 0.99(
). So we have T = 2.68
The margin of error is:
M = T*s = 2.68*2.4 = 6.432.
In which s is the standard deviation of teh sample. So
The lower end of the interval is the sample mean subtracted by M. So it is 23.6 - 6.432 = 17.168 minutes
The upper end of the interval is the sample mean added to M. So it is 23.6 + 6.432 = 30.032 minutes
The 99% confidence interval for the mean time required by all individuals to compete the form is between 17.168 minutes and 30.032 minutes.
Answer:
The percentage of her shots that will travel less than 180 yards is approximately 15.9%
Step-by-step explanation:
First it will be required to find the standard or z-score from the following formula
Where:
z = z or standard score
x = value that is observed = 180 yards
μ = Sample mean = 200 yards
σ = Sample standard deviation yards
Plugging the values, we have;
Therefore, the probability that her shots will travel less than 180 is given by the relation;
p( x < 180 yards) = p(z < -1) = 0.15866
Therefore, the percentage of her shots that will travel less than 180 = 0.15866 × 100
= 15.866% ≈ 15.9%.
Answer:
Yes the stand would make a profit, a $25 profit to be exact.
Step-by-step explanation:
the equation is 3n>100+0.50n
if the stand sells 50 then n=50 making it 3(50)>100+0.50(50).
by doing the multiplication then adding you'll get 150>125 since 3×50 is 150 ans 0.50×50 is 25 and you add the 25 to 100.
which means it'll gain $150 and spend $125 on maintaining the stand and for supplies with a $25 left over. It would make a profit
Answer:6842
Step-by-step explanation: 9378-2536=6842
