![\bf \textit{sum of all interior angles in a polygon}\\\\ S=180(n-2)~~ \begin{cases} n=\textit{number of sides}\\[-0.5em] \hrulefill\\ n=8 \end{cases}\implies S=180(8-2) \\\\\\ S=180(6)\implies S=1080](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bsum%20of%20all%20interior%20angles%20in%20a%20polygon%7D%5C%5C%5C%5C%20S%3D180%28n-2%29~~%20%5Cbegin%7Bcases%7D%20n%3D%5Ctextit%7Bnumber%20of%20sides%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D8%20%5Cend%7Bcases%7D%5Cimplies%20S%3D180%288-2%29%20%5C%5C%5C%5C%5C%5C%20S%3D180%286%29%5Cimplies%20S%3D1080)
now, we know 7 of those angles are already 835°, so the last one must be 1080 - 835 = 245°.
-2/5
Step by Step :
18 - 8 = 10
-15 - 10 = -25
10/-25 = -2/5
Answer:
Ummmm what’s the question?
Step-by-step explanation:
I will edit after you tell me what your question is.
Btw God bless you.have a great day!
<span>By the Pythagorean theorem
AB = </span>√(12²+5²) = √169 = 13 units
Answer:
The IQR is given by:

If we want to find any possible outliers we can use the following formulas for the limits:


And if we find the lower limt we got:


So then the left boundary for this case would be 3 days
Step-by-step explanation:
For this case we have the following 5 number summary from the data of 144 values:
Minimum: 9 days
Q1: 18 days
Median: 21 days
Q3: 28 days
Maximum: 56 days
The IQR is given by:

If we want to find any possible outliers we can use the following formulas for the limits:


And if we find the lower limt we got:


So then the left boundary for this case would be 3 days