Friday, July 02, 2021

Algebra Problems of the Day (Integrated Algebra Regents, January 2013)



While I'm waiting for new Regents exams to come along (AND THEY ARE COMING SOON!), I'm revisiting some of the older NY Regents exams.

More Regents problems.

Administered January 2013

Part IV: Each correct answer will receive 4 credits. Partial credit can be earned.


37. Using the line provided, construct a box-and-whisker plot for the 12 scores below.

26, 32, 19, 65, 57, 16, 28, 42, 40, 21, 38, 10

Determine the number of scores that lie above the 75th percentile.

Answer:


Before you do anything else, PUT THE NUMBERS IN ORDER!

There are 12 scores, so the median is between the sixth and seventh (which is the sixth from the back). That means Q1 is between the third and fourth, and Q3 is between the ninth and tenth. You don't need to "take the average", just find the number in between them. In each case, it's only 1 or 2 away. No need to "add them and divide by 2".

The minimum is the first (lowest) number, and the maximum is the last (highest) number. You should label all five of these numbers so that they are somewhere on the page. In particular, there is no obvious way from the number line to know that Q3 is 41.

Pick a scale for the number line. It is big enough to count by fives. You DO NOT have to start at 0. I did, but there isn't any particular reason to. You could have started with 5 or 10.

Do NOT label the number line with the scores. What I mean here: don't write 10, 16, 19, etc. under the tick marks. That's not how a number line works. And I've seen it written by students who didn't understand what they were supposed to do.

Since the third-quartile (Q3) is 41, there are THREE numbers greater than it. To get this credit, you need to have Q3 stated somewhere on the page, and you should identify those three numbers. You can write the three numbers, or where you wrote them in order, you can circle them and write that they are above the Q3.

The image below shows the box-and-whisker plot.





38. A metal pipe is used to hold up a 9-foot fence, as shown in the diagram below. The pipe makes an angle of 48° with the ground.

Determine, to the nearest foot, how far the bottom of the pipe is from the base of the fence.

Determine, to the nearest foot, the length of the metal pipe.

Answer:


You are given an angle and the length of the opposite side. That means that you have two trigonometric functions avaiable to you:
sin (48°) = opposite / hypotenuse
tan (48°) = opposite / adjacent

Use the sine function to find the length of the metal pipe, and the tangent function to find the distance on the ground. You will get half credit if you reverse these. You will lose points if you cosine in place of one of the other two functions.

Some common sense: the metal pipe MUST be longer than 9 feet because it's the hypotenuse. It you get a number smaller than 9, you made a mistake. Since the angle is greater than 45, that means that the adjacent side will be smaller than opposite side, but since it is 48 degrees, it will not be shorter by a lot.

Finally, make sure your calculator is in DEGREE mode not RADIAN mode or you will get crazy results. Conversely, if you get crazy results, check if your calculator is in RADIAN mode.

tan (48°) = 9 / x
x = 9 / tan (48°) = 8.1...
The bottom is approximately 8 feet from the fence.

sin (48°) = 9 / x
x = 9 / sin (48°) = 12.1...
The metal pipe is approximately 12 feet long.





39. On the set of axes below, graph the following system of equations.

y + 2x = x2 + 4
y - x = 4

Using the graph, determine and state the coordinates of all points in the solution set for the system of equations.

Answer:


You are graphing a parabola and a line. They may intersect at one point, two points or zero points. Note that even if you make graphing errors, you can get points for stating the solutions (even if you have to approximate a decimal) that are shown ON YOUR GRAPH. If your two lines don't intersect, then you must state that there are No Solutions to the system.

The first mistake that might be made usually happen when rewriting the equations to isolate the y on the left. This will allow you to use your graphing calculator to help you create a Table of Values.

y + 2x = x2 + 4
y = x2 - 2x + 4
This will have its Axis of Symmetry at x = 1.

y - x = 4
y = x + 4
Slope = 1, y-int = 4

Since the parabola is symmetrical, if you start with x = 1, 2, 3, 4, you will get the results for x = 0, -1, -2 as well.

As you can see in the image below, there are two points of intersection, (0, 4) and (3, 7). You can label these points on the graph. However, if you identify any other points on the graph, then you need to label these as your solutions.

Also, AT LEAST ONE of the two lines must be labeled with the original equation from the problem, not the rewritten one. Yes, it may be "obvious" to you which line is the quadratic and which is linear, but it must be stated. Not doing this is a graphing error and you will lose a point.




End of Part IV
End of Exam.


How'd you do?



More to come. Comments and questions welcome.

More Regents problems.

No comments:

Post a Comment