## Tuesday, June 18, 2013

### More on the Algebra Regents

First off, if you weren't familiar with the word bivariate, you could have broken it down into bi-, meaning "two", and -variate, which looks like "variable", right? So bivariate: two variables. Which of the tables is measuring two variables and will give a scatter plot, as opposed to a bar graph. The answers, unfortunately, doesn't matter because the question was thrown out. A "lack of specificity" was the reason.

Likewise, if you took the test in Chinese, two answers were accepted to question number 1 because of a translation error. That happens a lot.

As for the Factorial question, a.k.a. "the question with the exclamation point", I was able to guess that answer without doing any work for one simple reason: the last step was to subtract 10, but only one answer was 10 less than another. That answer was the correct one. (I checked my guess afterward, of course.) For the record: 6! + 5!(3!)/(4!) - 10 can be done with the scientific calculator, if you know where to look, but it isn't necessary.
6! = 720, 3! = 6, 5!/4! = 5, so 720 + 5 * 6 - 10 = 720 + 30 - 10 = 740

I wanted to review some of the open-ended questions.

Question 31. An inequality with a negative multiplier. The trick was to remember to reverse the direction of the inequality symbol.

That is, -5(x - 7) < 15, when divided by -5 becomes (x - 7) > -3.
The final answer is x > 4.

Question 32. A volume question on the Algebra test. Silly. If they at least gave the Volume and asked to find, say, the height, you could argue it was an Algebra task, but, as is, it's a middle-school problem.

The formula for volume of a cylinder was in the back of the booklet: V = (pi)r^2*h.
The trick here is that the gave the diameter instead of the radius, so you had to divide 13 by 2 to get 6.5. If you didn't, your answer was four times larger than it should've been, but you most likely got 1 out of 2 points.

The final answer is 1,014*pi. Note: The question said "in terms of [pi]", so if you multiplied by 3.14 or used the pi key on your calculator (i.e., you did extra work!), you lost a point for not answering the question that they asked.

Question 33. A distance question with big numbers, with a conversion added on. Two questions on the test involved converting between hours and days and weeks. This was one of them.

The distance from Earth to Mars is 136,000,000 miles. A spaceship travels at 31,000 miles per hour. Determine, to the nearest day, how long it will take the spaceship to reach Mars.

Divided 136,000,000 by 31,000 to get the number of hours (4387.096774...) and then divide by 24 to get the number of days (182.795698...).
The final answer is 183 days.

Question 34. The Counting Principle. How many options are on the menu? They've given this question many times before, but this is the largest number of items that they've ever used. The Principle remains the same.

There are five main courses, three vegetables, five desserts, and three beverages. To find the number of possible means, multiple the four of them: 5 * 3 * 5 * 3 = 225.
How many have chicken tenders? That's 1 * 3 * 5 * 3 = 45, which is one-fifth of the total.
How many have pizza (1), corn or carrots (2), a dessert (5) and a beverage (3): 1 * 2 * 5 * 3 = 30

If you showed your work, you likely got one point for each correct answer.

Question 35. Trigonometry. Find the angle of elevation.
You have a right triangle with a height of 350 feet and a base of 1000 feet, and you want to find the angle on the ground. You have the opposite (350) and the adjacent (1000), but you don't know the hypotenuse, so that means that you need to use tangent to solve the problem.

So tan(x) = (350)/(1000) and, therefore x = tan-1(350/1000), which is approximate 19.29.
The final answers is 19 degrees
Partial credit likely for using sine or cosine, or if you answer is expressed in radians or if rounded incorrectly.

Question 36. Summation of radicals. I cover this in Geometry. I have simplified them when dealing with Pythagorean Theorem but I don't usually cover this as it requires an extra day or so that I don't have.

Bear with me as I try to type this without any graphics:

(25)^.5 - 2(3)^.5 + (27)^.5 + 2(9)^.5
"^" means raise to a power, "^.5" means raise to 1/2 power, which means square root.

The square root of 25 is 5 and twice the square root of 9 is 2*3 = 6. That leaves the root 3 term, which is in simplest form, and root 27, which simplifies to (9*3)^.5 = 3(3)^.5. So 5 + 6 = 11 and -2(3)^.5 + 3(3)^.5 = 1(3)^.5.
The final answers is 11 + (3)^.5. In other words, 11 + the square root of 3.

Question 37. Algebraic fractions.

This is the one time where your teacher may have given you bad advice. Actually, it was good advice, if you know when to use it, but this isn't the time.
I have colleagues who will tell students (the ones who hate fractions or just "can't do" them) to multiply by the denominators to get rid of them. In this case, that would make a big mess. Don't do that.
This is one approach you could take:

2 / 3x + 4 / x = 7 / (x + 1)
2 / 3x + 12 / 3x = 7 / (x + 1)
14/ 3x = 7 / (x + 1)

At this point, cross-multiplying will yield the equation 14(x + 1) = 21x
The final answers is x = 2.

Question 38.Probability. I was expecting more of a twist for a four-point problem. This one isn't too bad if you know what you're doing.
Five red marbles and three green marbles make eight marbles total
These are dependent events, and their probabilities will be multiplied.
P(red then green) = 5/8 * 3/7 = 15 / 56
P(both red) = 5/8 * 4/7 = 20 / 56 = 5 / 14
P(both red or both green) = P(both red) + P(both green) = (20 / 56) from above plus 3/8 * 2/7 = 6 / 56. The total is 26/56, which reduces to 13/28.
I recommend to my students to reduce fractions, but I can't say if it's required for full credit in this problem.

Question 39. An Area problem, which I have never covered in Algebra.
You needed to find the area of the rectangle, then the area of the semicircle, and then subtract the latter from the former.
The tricky part here was understanding the frickin' question! Seriously, they went out of the way to be obtuse about it, making this more of a reading comprehension problem than an actual math problem.
In the diagram, AB = 5, and AB = BC = DE = FE, but CD = 6, which means the radius, which can't be named because they didn't name the point at the center of the circle, is 3.

The rectangle is 5 X 16, so the area is 80.
The semicircle is 1/2(pi)(3)^2 = 4.5(pi), which is approximately, 14.13716694.
The shaded area is 80 - 14.13716694 = 65.86283306, which rounds to 65.86.
The final answers is 65.86 square inches.

STUPID QUESTION ALERT! The folks who write these tests know very well that students have been taught for years to approximate pi as 3.14, whether or not they have a calculator which has a "pi" key on it. Using 3.14, the numbers will change in this problem. The semicircle will become 14.13, and the final answer 65.87. Whether or not a student loses a point on this may very well depend upon the teacher scoring it.