Which table represents y as a function of x?
1 answer:
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Segment NO is parallel to the segment KL.
Solution:
Given KLM is a triangle.
MN = NK and MO = OL
It clearly shows that NO is the mid-segment of ΔKLM.
By mid-segment theorem,
<em>The segment connecting two points of the triangle is parallel to the third side and is half of that side.</em>
⇒ NO || KL and
Therefore segment NO is parallel to the segment KL.
Check the picture below
now, <span>26°35' is just 26bdegrees and 35 minutes
your calculator most likely will have a button [ </span><span>° ' " ] to enter degrees and minutes and seconds
there are 60 minutes in 1 degree and 60 seconds in 1 minute
so.. you could also just convert the 35' to 35/60 degrees
so </span>![\bf 26^o35'\implies 26+\frac{35}{60}\implies \cfrac{1595}{60}\iff \cfrac{319}{12} \\\\\\ tan(26^o35')\iff tan\left[ \left( \cfrac{391}{12} \right)^o \right]](https://tex.z-dn.net/?f=%5Cbf%2026%5Eo35%27%5Cimplies%2026%2B%5Cfrac%7B35%7D%7B60%7D%5Cimplies%20%5Ccfrac%7B1595%7D%7B60%7D%5Ciff%20%5Ccfrac%7B319%7D%7B12%7D%0A%5C%5C%5C%5C%5C%5C%0Atan%2826%5Eo35%27%29%5Ciff%20tan%5Cleft%5B%20%5Cleft%28%20%5Ccfrac%7B391%7D%7B12%7D%20%5Cright%29%5Eo%20%5Cright%5D)
now, the angle is in degrees, thus, make sure your calculator is in Degree mode
now, <span>26°35' is just 26bdegrees and 35 minutes
your calculator most likely will have a button [ </span><span>° ' " ] to enter degrees and minutes and seconds
there are 60 minutes in 1 degree and 60 seconds in 1 minute
so.. you could also just convert the 35' to 35/60 degrees
so </span>
now, the angle is in degrees, thus, make sure your calculator is in Degree mode
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.