Answer: a. 1.981 < μ < 2.18
b. Yes.
Step-by-step explanation:
A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.
First, we calculate mean of the sample:
2.08
Now, we estimate standard deviation:
s = 0.1564
For t-score, we need to determine degree of freedom and 
:
df = 12 - 1
df = 11
= 1 - 0.95
α = 0.05
0.025
Then, t-score is
= 2.201
The interval will be
± 
2.08 ± 
2.08 ± 0.099
The 95% two-sided CI on the mean is 1.981 < μ < 2.18.
B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.
-4/5 = -16/20, so the new expression is (-16/20)+(3/20)
then, -16 + 3 is -13, so the solution is -13/20
Answer:
47 and the median
Step-by-step explanation:
Answer:
78
Step-by-step explanation:
The IQR is quartile 3 - quartile 1. First, we need to find the mean of the dataset, which is 50 -- the fifth number. Then, the median of the dataset on the left of the fifth number is the first quartile, and the median of the dataset on the right is the third. We get Q1 to then be halfway between 30 and 50, or 40, and Q3 to be halfway between 56 and n, so Q3 = (56+n)/2. Since we need Q3-Q1=27, or (56+n)/2-40=27, we have (56+n)/2=67 and 56+n=134, so n = 134-56=78
Part A: After 9 days the radius of algae was approximately 12.81 mm. The reasonable domain to point is (0,9).
Part B: The 9 on the y-intercept represents the amount of algae the experiment started with.
Part C: (12.81-10.12)/7=0.38
