well, 26+6 is 32, so divide that by 3 and you'd get 10.67, making daniel 32 by that time and his son being 10 years old
its simple math really, also, i hope this helps you ^-^
Answer:
The interval [32.6 cm, 45.8 cm]
Step-by-step explanation:
According with the <em>68–95–99.7 rule for the Normal distribution:</em> If 
is the mean of the distribution and s the standard deviation, around 68% of the data must fall in the interval
![\large [\bar x - s, \bar x +s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-%20s%2C%20%5Cbar%20x%20%2Bs%5D)
around 95% of the data must fall in the interval
![\large [\bar x -2s, \bar x +2s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-2s%2C%20%5Cbar%20x%20%2B2s%5D)
around 99.7% of the data must fall in the interval
![\large [\bar x -3s, \bar x +3s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-3s%2C%20%5Cbar%20x%20%2B3s%5D)
So, the range of lengths that covers almost all the data (99.7%) is the interval
[39.2 - 3*2.2, 39.2 + 3*2.2] = [32.6, 45.8]
<em>This means that if we measure the upper arm length of a male over 20 years old in the United States, the probability that the length is between 32.6 cm and 45.8 cm is 99.7%</em>
The coordinates of A will be (2P +M)/3
= (2(16, 14) +(1, 4))/3 = (33/3, 32/3) = (11, 32/3)
The appropriate choice is
(C) (11, 32/3)
_____
You will note that the coordinates of A are the weighted average of the coordinates of the end points. The weighting is the reverse of the ratio of the line segments. That is, the point adjacent to the shortest segment gets the highest weighting. (This is typical of the solution to "mixture" problems.)
