Answer:
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Step-by-step explanation:
Previous concepts
The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

Solution to the problem
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Answer:
Hello,
Step-by-step explanation:
![A=(1,2)\\B=(0,-1)\\\overrightarrow{AB}=((0,-1)-(1,2)=(-1,-3)\ ||\overrightarrow{AB}||^2=1+9=10\\\overrightarrow{BC}=((3,-2)-(0,-1)=(3,-1)\ ||\overrightarrow{BC}||^2=9+1=10\\\\Triangle\ is\ isosceles.\\\\\overrightarrow{AB}.\overrightarrow{BC}=(-1,-3)*\left[\begin{array}{c}3\\-1\end{array}\right] =-3+3=0\\\\Triangle \ is\ right.\\\\](https://tex.z-dn.net/?f=A%3D%281%2C2%29%5C%5CB%3D%280%2C-1%29%5C%5C%5Coverrightarrow%7BAB%7D%3D%28%280%2C-1%29-%281%2C2%29%3D%28-1%2C-3%29%5C%20%7C%7C%5Coverrightarrow%7BAB%7D%7C%7C%5E2%3D1%2B9%3D10%5C%5C%5Coverrightarrow%7BBC%7D%3D%28%283%2C-2%29-%280%2C-1%29%3D%283%2C-1%29%5C%20%7C%7C%5Coverrightarrow%7BBC%7D%7C%7C%5E2%3D9%2B1%3D10%5C%5C%5C%5CTriangle%5C%20is%5C%20isosceles.%5C%5C%5C%5C%5Coverrightarrow%7BAB%7D.%5Coverrightarrow%7BBC%7D%3D%28-1%2C-3%29%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%20%3D-3%2B3%3D0%5C%5C%5C%5CTriangle%20%5C%20is%5C%20right.%5C%5C%5C%5C)
Hi! to solve the question, how many necklaces did anne make in 16 hours if she makes 10 items in 2 standard 8-hour work days, we should first check the time she needs to make each item.
3 individual hours to make a necklace, and 1 hour to make a ring. according to the problem, she made 10 items in 16 hours. this problem will most likely be trial and error. let’s try one first guess.
my first guess will be, maybe she will make 6 rings and 4 necklaces. 6 times 1 is 6, which takes 6 hours for the rings. 4 times 3 is equal to 12. 6 + 12 is too high! let’s try again.
what about 6 rings and 2 necklaces? 6 times 1 is 6, so that’s 6 hours. 2 times 3 is 6 hours! 6 plus 6 is 12, which is close, but not quite there!
let’s try 7 rings and 3 necklaces! 7 times 1 is 7 hours. 3 times 3 is equal to 9 hours! total that together, 7 plus 9, is equal to 16 hours spent!
so, the answer would be, “Anne made 7 rings and 3 necklaces within 16 hours.” hope this helped!:)
This is a tricky question, all you need to know is that the remainder cannot be a negative number.
So, -50 mod 7 = 6, since -50 /7 = 8 with remainder of 6