You can make 6 different towers. If red was at the bottom, there'd be two different possibilities for that: red, yellow, blue, and red, blue, yellow. Same for if blue was at the bottom: blue, red, yellow, and blue, yellow, red. Yellow also applies: yellow, red, blue, and yellow, blue, red. 3*2=6, so 6 possible towers.
Answer:
16
Step-by-step explanation:
16+19=35
19-16=3
16 green peppers
You can calculate them out and check which one is the most accurate
so if one of the choices was 1/3 of 50, then you would do:
1/3 in the calculator (should be 0.333...) times 50
And you can compare it with the EXACT number, which is:
0.25*53=13.25
This way you can test all the options
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
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* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
Answer: Choice A) mean, there are no outliers
Have a look at the image attached below. I made two dotplots for the data points. The blue points represent bakery A. The red points represent bakery B. For any bakery, the points are fairly close together. There is no point that is off on its own. So there are no outliers, making the mean a good choice for the center. If there were outliers, then the median is a better choice. The mean is greatly affected by outliers, while the median is not.
