The prime factorisation of 600 is given by
Therefore, a = 3, b = 3, c = 5 and d = 2.
Important: Please use " ^ " to indicate exponentiation:
<span>"f(x) =x^2 to the number of x-intercepts in the graph of g(x) = x^2 +2."
Notes: the graph of f(x) = x^2 is a vertical parabola that opens up. It has its vertex at (0,0). This is the only point at which f(x)=x^2 has a horiz. intercept.
g(x) = x^2 + 2 has a graph that looks the same as that of f(x) = x^2, EXCEPT that the whole graph is moved 2 units UP. This new graph never touches or intersects the x-axis. Therefore, g(x) has NO horiz. intercepts (no x-int.).
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We would assume the lot takes a rectangular shape. With that in mind. the area of the rectangular plot is 48, 000 and its width is 240. We only need to know the area of a rectangle which is L * W.
So 48, 000 = L * 240
L = 48, 000/ 240
L = 200
It is 200 square feet deep.
Answer:
1. Take the Average of the distances the ball travelled each hit.
2. He should use the Interquartile Range. This is the difference between the Upper Quartile and the Lower Quartile of the distances he hits the ball.
3. He should use Mean
4. He should use Median. It best measures skewed data
Step-by-step explanation:
THE FIRST PART.
Raul should take the average of the distances the ball travelled each hit.
This is done by summing the total distances the ball travelled each bounce, and then dividing the resulting value by the total number of times he hit the ball, which is 10.
THE SECOND PART
He should use the Interquartile Range. This is the difference between the Upper Quartile and the Lower Quartile of the distances he hits the ball.
THE THIRD PART
He should take the mean of the distances of the ball that stayed infield.
This is the distance that occurred the most during the 9 bounces that stayed infield. The one that went outfield is makes it unfair to use any other measure of the center, taking the mean will give a value that is significantly below his efforts.
THE FOURTH PART
He should take the Median of the data, it is best for skewed data.
This is the middle value for all the distances he recorded.
