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.
2 is the GCF 3+8k
Hope this answer helps you
The 4th one, since both values have less than .5 after the decimal
Its D if you want to give it a shoot but im 75% sure is D hope you got it right. :)
Plane #1 speed: x mph
Plane #2 speed: (x+30) mph
time = time becomes
170 mi 185 mi
---------- = --------------
x x+30
Solving for x, the speed of the slower plane, we get
170x + 5100 = 185 x
5100
Then 15x = 5100, and x = ---------- (mi/hr) = 340 mph
15
The slower plane flies at 340 mph, and the faster one at 370 mph.