I am currently scoring exams, and while I cannot talk about that specifically, I do have a copy of the questions. The work you see below is mine, written on photocopies of the pages that they gave me for scoring purposes.
A cleaner, fuller (and possibly more clear) explanation of the problems may come at a future date. However, this is likely the last time that this test will be administered. Then again, they have said that before.
All questions in this section were open-ended and worth 3 points. A computational or rounding error cost 1 point. A conceptual error (e.g., using the wrong formula) cost 2 points. One of each cost all 3 points, and the answer -- despite the amount of work -- is worth 0. (There are exceptions, where the rubric states a point is awarded for finding a specific checkpoint in the middle of the problem and then not having any correct work afterward.)
34. Ryan bought three bags of mixed tulip bulbs at a local garden store. The first bag contained 7 yellow bulbs, 8 red bulbs, and 5 white bulbs. The second bag contained 3 yellow bulbs, 11 red bulbs, and 6 white bulbs. The third bag contained 13 yellow bulbs, 2 red bulbs and 5 white bulbs. Ryan combined the contents of these three bags into a single container. He randomly selected one bulb, planted it, and then randomly selected another and planted that one. Determine if it is more likely that Ryan planted a red bulb and then another red bulb, or planted a yellow bulb and then a white bulb. Justify your answer.
You needed to find the compound probabilities of two dependent event (without replacement) and compare the two.
There was no guideline in the rubric, and it didn't come up in our discussion, if you only found the number of outcomes and compared the two with the proper justification.
When added together, there were 23 yellow, 21 red and 16 white for a total of 60 bulbs. There would only be 59 when drawing the second bulb.
P(R then R) = (21 / 60) * (20 / 59) = 420/3540
P(Y then W) = (23 / 60) * (16 / 59) = 368/3540
Therefore, it is more likely that he would get red then red.
Note: some students did write these numbers are decimals or percents. Their math was checked for accuracy (no rounding needed), and then the comparison.
See image below:
35. A particular jewelry box is in the shape of a rectangular prism. The box is advertised as having an interior length of 20.3 centimeters, an interior width of 12.7 centimeters, and an interior height of 10.2 centimeters. However, when a customer measures the interior of the box, she finds that the interior height is actually 6.3 centimeters. Upon further examination, she discovers that the bottom of the interior of the box lifts up to reveal a hidden compartment. Find the volume of this hidden compartment to the nearest cubic centimeter.
Rounding matters. The right formula matters.
I was ready to go off on a rant (and I did, on Twitter) because so many students lost 2 of the 3 points because they used the formula for Surface Area. This meant that they had to do a lot more work -- and do it perfectly -- to get a single point. A rounding error and it would be zero.
Why did so many students do Surface Area? My guess: the formula is in the back of the book. Volume = Length X Width X Height is not, likely because it's so easy that every student should know it. But for whatever reasons, these students didn't.
The height of the hidden compartment was 10.2 - 6.3 = 3.9. Volume = 12.7 X 20.3 X 3.9 = 1005.459 = 1005 cm3.
Rounding before you multiplied was a 1-point error.
Alternatively, you could have calculated Volume = 12.7 X 20.3 X 10.2 = 2629.662 and Volume = 12.7 X 20.3 X 6.3 = 1624.283 and then subtracted to get the same answer. More work, but not incorrect.
See image below:
36. Solve algebraically for all values of x that satisfy the equation: x / (x + 4) = 3 / (x + 2)
Cross-multiply the numerators and denominators. Use the Distributive Property.
This gives you x2 + 2x = 3x + 12
Subtract everything from the right side to get: x2 - x - 12 = 0. (This was good enough for 1 point.)
Factor: (x - 4)(x + 3) = 0
x - 4 = 0 or x + 3 = 0
x = 4 or x = -3
Note: If you made a computational error, you probably had a problem factoring. However, if you used the Quadratic Formula and followed through to get an answer, however complicated it might have been, it would only be a 1-point error. We had a couple of these.
See image below:
How did you do?
If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.