9514 1404 393
Answer:
y -2 = 2(x -2)
Step-by-step explanation:
The slope of the line is given by the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (4 -2)/(3 -2) = 2/1 = 2
The point-slope form of the equation for the line is ...
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
For the first point and the slope we found, the equation is ...
y -2 = 2(x -2)
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You can rearrange this to any form you may like.
y -2 = 2x -4 . . . eliminate parentheses
y = 2x -2 . . . . . slope-intercept form
2x -y = 2 . . . . . standard form
Answer:
7. C(N(h))=33hx+460h
8. about $13.94 is the cost
Step-by-step explanation:
7. C(x)=(33x+460)(N(h))=(N(40)h)--C(x)=(33x+460)(40h)--C(x)=1320hx+18400h--simplified to C(N(h))=33hx+460h
8. C(N(10))=33(10)x+460(10)--C(N(10))=330x+4600--4600/330=about13.94
Answer:
Associative Property
Step-by-step explanation:
Answer: Choice C
Amy is correct because a nonlinear association could increase along the whole data set, while being steeper in some parts than others. The scatterplot could be linear or nonlinear.
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Explanation:
Just because the data points trend upward (as you go from left to right), it does not mean the data is linearly associated.
Consider a parabola that goes uphill, or an exponential curve that does the same. Both are nonlinear. If we have points close to or on these nonlinear curves, then we consider the scatterplot to have nonlinear association.
Also, you could have points randomly scattered about that don't fit either of those two functions, or any elementary math function your teacher has discussed so far, and yet the points could trend upward. If the points are not close to the same straight line, then we don't have linear association.
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In short, if the points all fall on the same line or close to it, then we have linear association. Otherwise, we have nonlinear association of some kind.
Joseph's claim that an increasing trend is not enough evidence to conclude the scatterplot is linear or not.
Answer
six dollars
Step-by-step explanation:
