a) First, draw a graph. The x axis should be numbered: 0, 1, 2, 3, 4, 5, 6, 7, 8. Each number is the last digit of the year. For each year, there is one average score that has to be plotted. Number the y axis of the graph from 0-600, by increments of 50, or something around there. Now, with each year (each x value) place the dot as high as the average score. Repeat with all years. DO NOT draw a line to connect them. A scatter plot is a bunch of points that are not connected.
b) Use the form (y2-y1)/(x2-x1) where (x1,y1) (x2,y2) are any two points from 2001-2006. It does not matter which points, pick any, and assign each of the numbers x1,y1 and x2,y2. Plug it in to the equation, and simplify. This is your slope, also called m. Now plug m into the equation y-y1=m(x-x1) where y1 and x1 are the x and y coordinates of any point from 2001 to 2006. x and y dont have values themselves, they stay there. Now distribute the m into the parenthesis, add y1 to both sides (to cancel it out on the left), and you should be left with the same equation, but now in y=mx+b form. B is where the line crosses the y axis, so put a point on the vertical axis where b is. M is your slope, so every time you go to the right one number, go up M numbers. Draw another point. Repeat this until you can connect these new dots into a line. This line should be on the same graph as your other points, but might not touch all of the scatter plot points. That's still okay, just leave it as is.
c) Find the point where x=6 (meaning the year is 2006) on your line. Find the average test score for that year by seeing where the point lines up along the y axis (draw a straight line from the point to the left to see, if you have to). Take that number and add 1m (the m is the number you used in step B) to it. This predicts what the average score might be after 1 (that's why the 1 is there) year. This is a predicted value, and might not be perfectly correct!! Write/record how close this new, predicted, score is to the actual number, 515, which can be found on the scatter plot.
d) repeat step C exactly as before (still use the year 2006) , but add 4m instead of 1m, to get the predicted score 4 years after 2006 (2010), instead of 1 year after (2007).
I hope I have been of some help! Best of luck!!! :)
Answer:
-4.65< -4 and 2/5
Step-by-step explanation:
To solve, this we can convert the fraction -4 and 2/5 into a decimal so that the values can be more easily compared. When we do this, we get the result that -4 and 2/5 equals -4.4. Now we can clearly see which value is greater because they are written in the same form:
-4.65<-4.4
-4.65<-4 2/5
Your answer would be ( A ) 132 inches
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
