## Thursday, August 14, 2014

### August 2014 Integrated Algebra Regents Exam, Open-Ended Questions

This morning was the Integrated Algebra Regents exam, and it was either the last one or next to last one. (There may be one final one in January.)

I've decided to start with the Open-Ended questions because they're quicker to type and I can scan in some of the answers I worked out.
There were three two-point questions, three three-point questions, and three fout-point questions. Partial credit was available for all of them, and work was required.

### Part 2

31. Using his data on annual deer population in a forest, Noj found the following information:

25th percentile: 12
50th percentile: 15
75th percentile: 22
Minimum population: 8
Maximum population: 27

Using the number line below, construct a box and whisker plot to display the data.

They gave you the Five-Number Summary. All you had to do was graph it. Start by using the number line, which had 30 hash marks, so you could've gone from 1 to 30, if you wanted to. You MUST have a scale designated, don't put one number here and one number there just because. Don't skip one here and skip two there.

32. The diagram below consists of a square with a side of 4 cm, a semicircle on the top, and an equilateral triangle on the bottom. Find the perimeter of the figure to the nearest tenth of a centimeter. [Diagram omitted.]

First of all, realize that if you didn't round correctly, you lost one point on a two-point problem. Don't throw away points!

The square has two sides showing, each side is 4. The equilateral triangle has two sides showing, and the third side is the side of the square, so those two sides are also 4. So far we have 4 + 4 + 4 + 4 or 16. Finally, the length of a semicircle (half the circumference) is 1/2 times pi times the diameter, which is the side of the square, which gives us 1/2 * 3.14 * 4, which is 6.28, rounded to 6.3, as per the question. Add 16 + 6.3 to get 22.3, which is the final answer.

NOTE: In this question, you could get away with using 3.14 for pi because the numbers were small and you had to round. That won't be the case in the next problem.

33. A thermos in the shape of a cylinder is filled to 1 inch from the top of the cylinder with coffee. The height of the cylinder is 12 inches, and its radius is 2.5 inches. State, to the nearest hundredth of a cubic inch, the volume in the thermos.

Note: If you used either 3.14 or 3.1416 for pi, you got the wrong answer. I am serious. If you didn't use the Pi key on the calculator, or at least use 3.141592, you answer will not round correctly. Will you lose points for this? You're supposed to, but it might slip by, depending on the scorer.

The formula for Volume is pi * r2 * h, where h is the height of the coffee, not the thermos, so h = 11, not 12.
The correct equation is V = (3.141592...)(2.5)2(11) = 215.98, to the nearest hundredth.

### PART 3.

34. The top of a lighthouse, T, is 215 feet above sea level, L, as shown in the diagram below. The angle of depression from the top of the lighthouse to a boat, B, at sea is 26o. Determine, to the nearest foot, the horizontal distance, x, from the boat to the base of the lighthouse.

The angle of depression from the top of the lighthouse is equal to the angle of elevation from the boat. We have a right triangle, and we know the side opposite angle B and we want the side adjacent to angle B. This means that we need to use tangent to solve the problem.

Tan 26 = opp/adj = 215/x. So x = 215/tan 26 = 440.815..., or 441, to the nearest foot.

35. There are six apples, five oranges and one pear in John's basket. His friend takes three pieces of fruit at random without replacement. Determine the probability that all three fruits taken are apples.

There are twelve pieces of fruit to begin with, six of which are apples. Each time he takes one, there is one fewer piece of fruit and one fewer apple.
The equation you needed to write was P = 6/12 * 5/11 * 4/10, which after cancelling factors (or multiplying and simplifying) leaves you with 1/11.

36. Express [THE FOLLOWING in simplest radical form.

I'll leave this for pictures as there are too many radical signs.

The y is there for no apparent reason, and the two radicals cannot be combined because they aren't alike. This was a bit of an unusual question, with three twists: the variable, the distribution of the negative sign and the different radicals in the answer.

### PART 4.

37. On the set of axes below, solve the following system of inequalities graphically:

y + 3 < 2x
-2y < 6x - 10
State the coordinates for a point in the solution set.

Again, I'll let the scan speak for me. Your answer should look similar if you rewrote the inequalities in slope-intercept form correctly. The "Ex", means "Example" and that was there for me as a scorer, as a reminder that there wasn't a single answer for that point.

Remember to label at least one of the inequalities. You can choose any point in the solution set, but don't pick a point on the broken line because that isn't part of the solution. It's a boundary.

38. The actual side of a square tile is 4 inches. The manufacturers allow a relative error of 0.025 in the area of a tile. Two machines are used to cut the tiles. Machine A produces a square tile with a length of 3.97 inches. Machine B produces a square tile with a length of 4.12 inches. Determine which machine produces a tile whose area falls within the allowed relative error.

This was another unusual question. It was easy in that they didn't have you calculate any surface areas or volumes. On the other hand, you had to calculate the relative error for both machines. This was also, arguably, a dumb question because the question implies strongly that the answer is Machine A. For this reason, if you said Machine A without any calculations, you will not get any points.

The work is shown in the scan below:

39. Solve the following system of equations algebraically:

y = x2 - 6x + 9
y = -9x + 19

Another scan. You didn't need to check your answers, but when I saw (-5, 64), I thought that that couldn't possibly be right, so I checked -- and it was. The other answer was a more reasonable (2, 1). You needed two find both x values and the y values associated with them.

And those are the open-ended questions.

Thanks for reading. I do this for you.

Danielle said...

I just wanted to leave a note thanking you for all of your effort in going through these exams. My daughter passed both the CC and Regents exams in June, but wanted to retake them to earn a better score. It has been 20+ years since I took an algebra class, but your website helped us tremendously as we reviewed this summer. Still waiting for the scores, but she feels confident that her grades will be higher this time around!

(x, why?) said...

Glad to hear it. Nice to know that these reviews are having an impact on students and parents.

Fingers crossed for good scores.

eve said...

I got a 98 on that test and thx for what i got wrong and right

(x, why?) said...

You're welcome, Eve, and congratulations on an excellent score!

Andy SSP said...

Hi I'd like to ask for question 39, why is 5 and -2 for the x value incorrect?

(x, why?) said...

X +5 = 0, so subtract 5 from both sides and you get x = -5.

Likewise for x - 2 = 0. Add 2 to both sides.