1012 divided by 120 is 8.43 so 9 boxes will be needed
Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
The 4 in the thousands place is 10 times the 4 in the hundreds place.
Answer: D. minimizes the sum of the squared residuals
Step-by-step explanation: The ordinary least square method is often used in locating the trendine which best fits a graphical linear model. The best is one in which the sum of the squared residual is smallest. The residual refers to the difference between the actual and the predicted points. The sum of the squared differences is obtained and the trend line is positioned where the residual is minimum. Choosing a OLS, and minimizing the sum.of the squared residual, the error difference between the predicted and actual score is minimized or reduced, hence, improving the prediction accuracy of our model.
