Answer:
1.
Mean: 65
MAD: 6.4
2.
Mean: 60.5
MAD: 6.4
3. The diference of the means are 4.5
I’m sorry, I do not understand the last one.
Step-by-step explanation:
The mean is the average, add all of the values and divide the result by the number of values. The MAD is the mean absolute deviation. You find it by taking the distance between the values and the mean. You then add them together and divide it by the mean. I really hope this helps you!
Answer:
The solutions are not viable because the amount of the sale cannot be negative
Step-by-step explanation:
We are told that;
Let A be the amount of the sale in dollars and C be the charge of the online auction
This means that both values of A and B should be non-negative integers.
Now, we are also told that An online auction charges sellers a flat fee of $1.20 and 2% of the amount of the sale.
We can conclude that the solution is not viable because the amount of sale of -1.1 cannot be negative.
I think equations because you have to know how to use numbers
Check the picture below.
![\stackrel{\textit{\Large Areas}}{\stackrel{triangle}{\cfrac{1}{2}(6)(6)}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}\pi (3)^2}}\implies \boxed{18+4.5\pi} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{pythagorean~theorem}{CA^2 = AB^2 + BC^2\implies} CA=\sqrt{AB^2 + BC^2} \\\\\\ CA=\sqrt{6^2+6^2}\implies CA=\sqrt{6^2(1+1)}\implies CA=6\sqrt{2} \\\\\\ \stackrel{\textit{\Large Perimeters}}{\stackrel{triangle}{(6+6\sqrt{2})}~~ + ~~\stackrel{semi-circle}{\cfrac{1}{2}2\pi (3)}}\implies \boxed{6+6\sqrt{2}+3\pi}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%286%29%286%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20%283%29%5E2%7D%7D%5Cimplies%20%5Cboxed%7B18%2B4.5%5Cpi%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bpythagorean~theorem%7D%7BCA%5E2%20%3D%20AB%5E2%20%2B%20BC%5E2%5Cimplies%7D%20CA%3D%5Csqrt%7BAB%5E2%20%2B%20BC%5E2%7D%20%5C%5C%5C%5C%5C%5C%20CA%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B6%5E2%281%2B1%29%7D%5Cimplies%20CA%3D6%5Csqrt%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Perimeters%7D%7D%7B%5Cstackrel%7Btriangle%7D%7B%286%2B6%5Csqrt%7B2%7D%29%7D~~%20%2B%20~~%5Cstackrel%7Bsemi-circle%7D%7B%5Ccfrac%7B1%7D%7B2%7D2%5Cpi%20%283%29%7D%7D%5Cimplies%20%5Cboxed%7B6%2B6%5Csqrt%7B2%7D%2B3%5Cpi%7D)
notice that for the perimeter we didn't include the segment BC, because the perimeter of a figure is simply the outer borders.
<h3>
Answer: x-2</h3>
Explanation:
If x > 2, then x-2 > 0 after subtracting 2 from both sides.
Since x-2 is always positive when x > 2, this means the absolute value bars around the x-2 aren't needed. The results of |x-2| and x-2 are perfectly identical.
For example, if we tried something like x = 5, then
- x-2 = 5-2 = 3
- |x-2| = |5-2| = |3| = 3
Both outcomes are 3. I'll let you try other x inputs.
So because |x-2| and x-2 are identical, this means |x-2| = x-2 for all x > 2.
In short, we just erase the absolute value bars.
