Answer:
The y-intercept of the line would be (C) 8
Step-by-step explanation:
The slope of the line with the equation y = -3x - 5 is -3. When figuring out a slope perpendicular to another line you would use the negative reciprocal of the slope which would lead to the slope of the line to be 1/3.
We know it passes through the point (-3,7), substituting x and y, we would then get our new equation,
.
Solve the equation by multiplying -3 with 1/3 and the product would be -1. Add negative one on both sides and you would get 8 = b.
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.
It is true because
when you separate 2y^2 +9y -18, you are supposed to get the factors
(2y-3)(y+6), when you multiply this it gives you the first equation, we know
this because when we multiply the first term it gives us exactly 2y^2 and if we
multiply the last ones it gives you -18, now here is the tricky part, when we
multiply the middle part we get 12y and -3y, when we add these up we get 9y,
now is that the same middle as in the equation? Yes… then our factors are correct
making this true
Hope this helps