Answer:
independent
Step-by-step explanation:
Dependent equations graph as the <em>same line</em>. These lines are not the same, so the equations are independent.
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However, the lines are parallel, so the equations are also <em>inconsistent</em>. There is <em>no solution</em> to this system of equations.
Part A:<span> What is the mean of the data? Show your work. (4 points)</span>
Part B:<span> Use your answer from Part A to calculate the mean absolute deviation for the data. Show your work. (6 points)</span>
9514 1404 393
Answer:
A, C
Step-by-step explanation:
The attached graph shows which lines go through the given point. They are ...
y = 1/2x -1 . . . . 1st selection
y = -1/6x +3 . . . 3rd selection
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The equations can be found algebraically by substituting the given point in the equation and seeing if the result is a true statement.
a) 2 = (1/2)(6) -1 = 3 -1 . . . true
b) 2 = -3(6) . . . . false
c) 2 = -1/6(6) +3 = -1 +3 . . . true
d) 2 = 2/3(6) -1 = 4 -1 . . . . false
e) 2 = 4(6) -2 = 24 -2 . . . . false
f) 2 = -3/2(6) +6 = -9 +6 . . . . false
Answer:
10
Step-by-step explanation:
We can set it up like this, where <em>s </em>is the speed of the canoeist:

To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
![s(s-5)[\frac{18}{s} + \frac{4}{s-5} = 3] \\ 18(s-5)+4s=3s(s-5) \\ 18s - 90+4s=3 s^{2} -15s](https://tex.z-dn.net/?f=s%28s-5%29%5B%5Cfrac%7B18%7D%7Bs%7D%20%2B%20%5Cfrac%7B4%7D%7Bs-5%7D%20%3D%203%5D%20%5C%5C%2018%28s-5%29%2B4s%3D3s%28s-5%29%20%5C%5C%2018s%20-%2090%2B4s%3D3%20s%5E%7B2%7D%20-15s)
If we rearrange this, we can turn it into a quadratic equation and factor:

Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.

The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.