Friday, May 31, 2019

SBAC Practice Test Questions 33-34 (Day 177)

Today I subbed in a high school math class. I'm not sure whether it's really worth doing "A Day in the Life," since all classes are taking a quiz today. Well, I might as well do it, since it might end up being the last math class I sub this school year.

7:05 -- The day begins with a "first period class," which really means zero period.

Most of the classes are our favorite math class -- Geometry. But today's quiz is on probability. As I've mentioned on the blog before, statistics and probability are included in the Common Core Standards, but it's up to the states to decide how to divide the standards into high school courses. Here in California, statistics is part of Algebra I and probability is part of Geometry.

Indeed, I once wondered whether I should follow the California route on the blog and end the year with probability instead of our SBAC Prep. After all, Geometry classes (including today's) typically have mostly sophomores, with only a few freshmen and (SBAC-taking) juniors. So our Geometry need more lessons on probability than on SBAC Prep.

As the test ends, I consider singing a song on probability -- Square One TV's "Ghost of a Chance" -- but I don't. (By the way, last week on the day the other district was closed, I subbed in a seventh grade class that was also studying probability, and I did sing "Ghost of a Chance" that day.)

7:55 -- First period leaves and second period arrives. These students take the quiz, but one girl doesn't have enough time to finish the quiz. So of course I can't sing any song.

8:45 -- Second period leaves and third period arrives. These students take the quiz, and this time there's enough time for me to sing "Ghost of a Chance."

But I'm hoping to sing a second song from that show -- "Combo Jombo" -- since at least one of the questions is on combinatorics. Unfortunately, I've never written down the lyrics, and Barry Carter has the lyrics to some Square One TV songs but not the combinatorics song. And so I can't sing it.

9:40 -- Third period leaves for snack.

10:00 -- Fourth period arrives. These students take the quiz, but one girl doesn't have enough time to finish the quiz. So of course I can't sing any song.

10:45 -- Fourth period leaves. It's now time for an assembly, which is all about the arrival and distribution of yearbooks. Teachers are to lead their fifth period classes to the gym for the assembly.

But fifth period happens to be my regular teacher's conference period. Teachers with fifth period conference are supposed to supervise the outside of the assembly instead, to make sure that students aren't trying to ditch school during the assembly.

11:45 -- The assembly ends and fifth period proper begins. It is now my conference period, which extends into lunch.

1:30 -- Sixth period arrives. This is the only Algebra I class of the day.

These students are also taking a test, on the Quadratic Formula. One or two students enter the room already singing Quadratic Weasel, so of course I sing the song with them before the exam as part of the test review.

2:15 -- Sixth period leaves. Teachers with first period typically don't have a seventh period -- and this includes my regular teacher. And so my day ends here.

There's not much to say on classroom management, since it's all just making sure the students are quiet -- and not using cell phones until all students have finished the quiz.

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

In the figure above the side opposite the _____-degree angle is the shortest.

All that's given is an unlabeled diagram, so let me label it. There are three non-overlapping triangles, ABC, ACD, and ADE, with the following angle measures:

Triangle ABC: BAC = 80, B = 22, ACB = 78
Triangle ACD: CAD = 70, ACD = 28, ADC = 82
Triangle ADE: DAE = 79, ADE = 31, E = 70

Of course, we don't find this answer by attempting to measure anything, since the diagram is clearly not to scale. (For example, the three angles at A add up to more than 180 degrees, yet on the diagram they appear to add up to less than 180.)

Instead, we are to use the Unequal Angles Theorem of Lesson 13-7:

Unequal Angles Theorem:
If two angles of a triangle are not congruent, then the sides opposite them are not congruent, and the longer side is opposite the larger angle.

We notice that in Triangle ABC, the side opposite 22 degrees (AC) is the shorter than any other side of that triangle. But it's not the shortest side of the entire figure. That's because in Triangle ACD, the side opposite 29 degrees (AD) is shorter than any other side of that triangle, including AC. But it's still not the shortest side of the entire figure. That's because in Triangle ADE, the side opposite 31 degrees (AE) is shorter than any other side of that triangle, including AD.

Thus AE is the shortest side of all, and this is the side opposite 31. Therefore the desired angle is 31 degrees -- and of course, today's date is the 31st.

Pappas gives this sort of question on her calendar at least once per year. She will always ask the question such that an answer can always be determined -- that is, it will always involve the sides common to two triangles (AC and AD above). For example, she won't ask for the longest side of the figure because we can't find it. The longest side of Triangle ADE is DE (opposite the 79) and the longest side of Triangle ACD is AC (opposite the 82). So we know both DE > AD and AC > AD, but there's no way to compare AC and DE -- without letting one side be 1 and then setting it up with the Law of Sines. If 70 and 79 were reversed, the longest side can be found without trig.

Question 33 of the SBAC Practice Exam is on writing an equation:

Mike earns $6.50 per hour plus 4% of his sales.

Enter an equation for Mike's total earnings, E, when he works x hours and has a total of y sales, in dollars.

Well, the equation almost writes itself. Just read the first sentence out loud:

Mike earns (E =) $6.50 (6.5) per hour (x) plus (+) 4% (0.04) of his sales (y).
E = 6.5x + 0.04y

The tricky part of course is writing the decimals 6.5 and 0.04 properly -- especially the conversion of 4% to a decimal.

Both the girl and the guy from the Pre-Calc class write the correct equation -- with my help, during class that day.

Question 34 of the SBAC Practice Exam is on writing and solving a system of equations:

The basketball team sold t-shirts and hats as a fund raiser. They sold a total of 23 items and made a profit of $246. They made a profit of $10 for every t-shirt they sold and $12 for every hat they sold.

Determine the number of t-shirts and the number of hats the basketball team sold.
Enter the number of t-shirts in the first response box.
Enter the number of hats in the second response box.

If we let s be the number of shirts and h be the number of hats. Then the equations are:

s + h = 23
10s + 12h = 246

Let's solve this system by substitution:

s + h = 23
s = 23 - h

10s + 12h = 246
10(23 - h) + 12h = 246
230 - 10h + 12h = 246
230 + 2h = 246
2h = 16
h = 8

s + h = 23
s + 8 = 23
s = 15

Therefore the team sold 15 t-shirts and eight hats.

The girl correctly writes and solves the system by substitution as I show above. Unfortunately, the guy leaves this problem blank.

In fact, throughout these 34 problems, it appears that the guy more easily gives up and avoids trying to work out the more difficult problems. The girl is at least more willing to attempt them.

At first, it appeared that one of the guy's strengths is factoring, as he did well on the early problems that require him to factor. But yesterday, the girl factored the quadratic function correctly while the guy mixed up two of the terms.

Both students, of course, have already long completed the SBAC. I wonder how well these two students fared on the state test, but I most likely will never know.

Today is our last activity day -- and since there is very little Geometry in either of today's problems, let me add some Geometry to the activities.

One of the activities is based on today's Pappas problem. I decide to interchange two of the angles as I mentioned above -- now Angle E = 79 and DAE = 70. This allows both the shortest and the longest sides to be determined.

The other activity is a probability problem -- since here in California, probability must be included in the Common Core Geometry course. It's actually Question 7 of the quiz from today's subbing.

Indeed, I decide that I will also post Questions 8-12 from both of today's assessments. (Neither one is under copyright from a textbook.) The Geometry quiz starts with the aforementioned combinatorics question followed by some deck of cards probability questions. The Algebra I test includes the Quadratic Formula and completing the square.

SBAC Practice Exam Question 33
Common Core Standard:
CCSS.MATH.CONTENT.HSA.CED.A.2
Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

SBAC Practice Exam Question 32
Common Core Standard:
CCSS.MATH.CONTENT.HSA.CED.A.3
Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.

Commentary: Both of these word problems should be straightforward, but the first is tricky for students who don't remember percents. Students are likely to give up on the conversion step and never reach solving the system for the last problem.





Thursday, May 30, 2019

SBAC Practice Test Questions 31-32 (Day 176)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

If the volume of the cube is 8 cubic cm, then half the measure of angle BAC is _____ degrees.

[Here is the given info from the diagram: A, B, and C are vertices of the cube, chosen so that segments AB, AC, and BC are all face diagonals.]

Like yesterday, we will let h be the height of the cube -- that is, it's the length of any edge. Then we can let d be the length of a face diagonal.

Thus triangle ABC has three sides all of length d -- in other words, it's equilateral. All equilateral triangles are equiangular -- hence angle BAC is 60. Therefore half of this measure is 30 degrees -- and of course, today's date is the thirtieth.

Yes -- we could have easily determined that h = 2 and d = 2sqrt(2). I choose not to do so above in order to emphasize that the answer would have been 30 regardless of the size of the cube. Indeed, just like yesterday, the fact that the triangle is equilateral turns out to be key.

Question 31 of the SBAC Practice Exam is on right triangle trigonometry:

Consider this right triangle.

[The right angle is at S, ST = 21, and RT = 35.]

Determine whether each expression can be used to find the length of side RS. Select Yes or No for each expression.

                 Yes  No
35 sin(R)
21 tan(T)
35 cos(R)
21 tan(R)

Let's first look at the two involving 35 times something R. We notice that 35 is the hypotenuse and relative to angle R, the desired side RS is the adjacent side, so we need the cosine. Thus 35 cos(R) is yes, while 35 sin(R) is wrong.

Now we check out the 21 times tangent something. We see that the desired side RS is opposite angle T while 21 is adjacent to it. So the tangent of T is RS/21. Thus 21 tan(T) is yes, while 21 tan(R) is no.

Both the girl and the guy from the Pre-Calc class answer the first three parts correctly. But the guy leaves the last part blank while the girl answers it correctly. Since the guy knows that 21 tan(T) is yes, it's most likely an accidental omission on his part.

Question 32 of the SBAC Practice Exam is on the graphs of quadratic functions:

Given the function
y = 3x^2 - 12x + 9,
  • Place a point on the coordinate grid to show each x-intercept of the function.
  • Place a point on the coordinate grid to show the minimum value of the function.
To find the x-intercepts of this parabola, let's factor the function:

y = 3x^2 - 12x + 9
y = 3(x^2 - 4x + 3)
y = 3(x - 1)(x - 3)

So the x-intercepts are at (1, 0) and (3, 0). Since the function vanishes when x is 1 or 3, it follows that the axis of symmetry, hence the vertex, is at x = 2. There are other ways to find the vertex, but it's best just to take the mean of the x-intercepts if we've already found them.

Let's now find the y-value of the vertex:

y = 3x^2 - 12x + 9
y = 3(2)^2 - 12(2) + 9
y = 12 - 24 + 9
y = -3

Therefore the vertex is at (2, -3). The SBAC only requires students to plot these three points -- once again, I'm not sure whether the full parabola can be graphed on the computer interface.

The girl from the Pre-Calc class correctly graphs these three points and tries to graph a parabola, even though her graph looks more like a V-shape. But the guy, unfortunately, makes an error in factoring:

y = 3x^2 - 12x + 9
y = 3(x^2 - 4x + 3)
y = 3(x - 4)(x + 1)

and so his x-intercepts are at (4, 0) and (-1, 0). In other words, he graphs y = 3(x^2 -3x - 4) instead of the correct graph. This counts as both a sign error as well as confusing the needed sum and product during factoring.

So what does the guy do for his vertex? For the x-value, it appears that he wants to make his parabola look symmetrical. The mean of his two x-intercepts is 1.5. But the vertex he draws ends up being closer to x = 2, which is unwittingly the correct value. He seems to choose a random value of y -- his vertex is at (2, -4), one unit below the correct vertex of (2, -3).

Ironically, the guy's graph actually looks more like a parabola than the girl's graph. But the girl is the one who correctly find the intercepts and minimum.

SBAC Practice Exam Question 31
Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.C.8
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

SBAC Practice Exam Question 32
Common Core Standard:
CCSS.MATH.CONTENT.HSF.IF.C.7.A
Graph linear and quadratic functions and show intercepts, maxima, and minima.

Commentary: The trig problem should be straightforward provided that the students know the definitions of the trig ratios. The parabola graphing will be tricky for the students who have trouble factoring the quadratic function.



Wednesday, May 29, 2019

SBAC Practice Test Questions 29-30 (Day 175)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

The lateral faces of this right prism are congruent squares. If the prism's volume is 10,560.75 cubic inches, what is its height to the nearest inch?

[Here is the given info from the diagram: it's a triangular prism.]

Notice that nowhere in this problem is it directly stated that the bases are equilateral triangles, but we can show that they are indeed equilateral. Here's how -- the question asks for the height of this prism, which we can call h. Then each side of the square lateral faces must be h. But each side of the triangular bases is also a side of a square, hence it must also be h.

So we must find the area of the base -- an equilateral triangle of side h. We see that the equilateral triangle can be divided into two 30-60-90 triangles with hypotenuse h and shorter leg h/2, and so the longer leg -- the altitude of the equilateral triangle -- must be h sqrt(3)/2. Thus the area of the equilateral triangle must be h^2 sqrt(3)/4.

Now we find the volume of the prism:

V = Bh
V = (h)h^2 sqrt(3)/4
V = h^3 sqrt(3)/4 = 10560.75

This is the only way to solve the problem -- we let the desired height be h, find the volume in terms of h, and then solve for h.

h^3 sqrt(3)/4 = 10560.75
h^3 = 10560.75(4)/sqrt(3)
h^3 = 24389.007
h = cbrt(24389.007)
h = 29.000003 inches

Therefore the desired height is 29" -- and of course, today's date is the 29th. Volumes of prisms appear in Lesson 10-5 of the U of Chicago text, but the equilateral triangle area needed to complete the problem can't be completed until special right triangles are studied in Lesson 14-1.

Let's get back to SBAC Prep/Final Exam Prep. Last year I never did cover the last six of the Released Test Questions. I fix this error by finally covering them now.

Question 29 of the SBAC Practice Exam is on angles of elevation:

Emma is standing 10 feet away from the base of a tree and tries to measure the angle of elevation to the top. She is unable to get an accurate measurement, but determines that the angle of elevation is between 55 and 75 degrees.

Decide whether each value given in the table is a reasonable estimate for the tree height. Select Reasonable or Not Reasonable for each height.

                Reasonable  Not Reasonable
4.2 feet
14.7 feet
24.4 feet
33.9 feet
39.1 feet
58.7 feet

Once again, let's use h for the height again. Angle of elevation problems usually depend on the tangent ratio, where h is the height and 10 is the distance to the tree:

tan theta = h/10

But we don't know what the angle of elevation theta is, except that it's between 55 and 75. So let's try solving the problem for both of the extreme values:

tan 55 = h/10
h = 10 tan 55
h = 14.3 feet

tan 75 = h/10
h = 10 tan 75
h = 37.3 feet

And the true height of the tree can be anywhere in between. We thus choose Reasonable for Emma's three heights in this range -- 14.7, 24.4, and 33.9 feet -- and Not Reasonable for her other three -- 4.2, 39.1, and 58.7 feet.

Both the girl and the guy from the Pre-Calc class correctly answer this question. Both of them draw right triangles to help them. The girl writes her work for 55 degrees, but not 75 degrees, while the guy starts to use sine, then corrects himself to tangent. Most likely, both of them enter 10 tan 55 on their calculators, so it was obvious that they needed to enter 10 tan 75 without writing out the work.

Question 30 of the SBAC Practice Exam is on modeling with linear equations:

Emily has a gift certificate for $10 to use at an online store. She can purchase songs for $1 each or episodes of TV shows for $3 each. She wants to spend exactly $10.

Part A
Create an equation to show the relationship between the number of songs, x, Emily can purchase and the number of episodes of TV shows, y, she can purchase.

Part B
Use the Add Point tool to plot all possible combinations of songs and TV shows Emily can purchase.

Since each song is $1 and each episode is $3, it's clear that the equation is 1x + 3y = 10. Notice that I include the coefficient for x even though it is 1, because the SBAC interface for Part A requires a coefficient for both variables.

For Part B, there are four possible solutions -- (1, 3), (4, 2), (7, 1), and (10, 0). These solutions are discrete, but I suspect that the SBAC interface automatically connects the points to form a line -- the graph of the linear equation 1x + 3y = 10.

Both the girl and the guy from the Pre-Calc class correctly answer Part A. But the guy's graph isn't linear, because he miscounts and graphs (6, 1) instead of (7, 1). The girl's graph is linear. But both of them miss the solution (10, 0) -- which is valid, as Emily could have bought 10 songs and no shows.

SBAC Practice Exam Question 29
Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.C.8
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

SBAC Practice Exam Question 30
Common Core Standard:
CCSS.MATH.CONTENT.HSA.CED.A.2
Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

Commentary: Both the tangent ratio and angles of elevation appear in Lesson 14-3 of the U of Chicago Geometry text. Meanwhile, Lesson 8-8 of the U of Chicago Algebra I text is called "Equations for All Lines." In that lesson, linear equations in standard form Ax + By = C are given, and it's stated that lines in standard form often arise naturally from real situations.



Tuesday, May 28, 2019

SBAC Practice Test Questions 27-28 (Day 174)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

Find the area of the octagon formed from this 6 * 6 square. The vertices of the hexagon are on points dividing the side in thirds.
[emphasis mine]

I emphasized two words above to highlight an error in this question -- first Pappas asks for the area of an octagon, but then she describes the vertices of a hexagon. Well, the square has four sides and it takes two points to trisect each side, and so there are eight vertices. Thus the intended polygon is clearly an octagon. And I can see the diagram -- and yes, it's definitely an octagon. Let's forgive Pappas for the typo and fix it ourselves:

Find the area of the octagon formed from this 6 * 6 square. The vertices of the octagon are on points dividing the side in thirds.

Well, the area of the square is 36, and one-third of the side length is two. Thus we have a composite figure formed by cutting off an isosceles right triangle of leg 2 from each vertex of the square. The area of each of the four cut-off triangles must be half of two times two, which is 2. So we have:

36 - 4(2) = 28

Therefore the desired area of the octagon is 28 square units -- and of course, today's date is the 28th.

As it turns out, there is a Geometry problem on the Pappas calendar everyday this week. So expect me to describe the Pappas problem in all of this week's blog entries.

Meanwhile, this is our final week of SBAC Practice Test questions. Surely, by now all high schools in both of my districts have completed the SBAC. My intention is for all SBAC review questions after the state test should become finals review questions instead.

Indeed, these review questions plus this week's Pappas problems together make up a nice review week for the final exam. Today's Pappas problem, for example, reviews Chapter 8 on area, especially Lessons 8-3 and 8-5.

This is what I wrote last year about today's lesson:

Question 27 of the SBAC Practice Exam is on quadrilaterals:

In the given figure, quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.

Ted claims that the two shaded triangles [ABC and DCE] are congruent. Is Ted's claim correct? Include all work and/or reasoning necessary to either prove the two triangles congruent or to disprove Ted's claim.

This is a Geometry question. Let's try to come up with a two-column proof of Ted's claim:

Given: Quadrilateral ABCD is a rectangle, and quadrilateral ACED is a parallelogram.
Prove: Triangle ABC = DCE.

Proof:
Statements                         Reasons
1. bla, bla, bla                    1. Given
2. ABCD is a pgram.         2. A rectangle is a parallelogram.
3. AB = DCAC = DE       3. Opposite sides of a pgram are congruent.
4. AB | | DCAC | | DE      4. Opposite sides of a pgram are parallel.
5. Angle BAC = ACD,       5. Alternate Interior Angles (parallel line consequences)
    Angle ACD = CDE
6. Angle BAC = CDE        6. Transitive Property of Congruence
7. Triangle ABC = DCE    7. SAS Congruence Theorem [steps 3,6,3]

This proof is sound, but it's tricky to enter a two-column proof into the SBAC. Most likely, the SBAC expects students to enter a paragraph proof into the box. We might try the following:

Paragraph Proof:
Since ABCD is a rectangle, it's also a parallelogram. Since opposite sides of a parallelogram are congruent, we have AB = DC and AC = DE. Since opposite sides of a parallelogram are also parallel by definition, AB | | DC and AC | | DE. This implies that alternate interior angles BAC and ACD are congruent, as are ACD and CDE. By the Transitive Property of Congruence, BAC = CDE. Therefore,
Triangles ABC and DCE are congruent by SAS. QED

It's interesting to think of this problem in terms of transformations. Notice that if a diagonal divides a parallelogram into two triangles, then a 180-degree rotation about the midpoint of the diagonal maps one triangle to the other. Thus a 180-degree rotation about the midpoint of AC maps Triangle ABC to CDA, and likewise a half-turn about the midpoint of CD maps CDA to DCE. So the composite of two isometries (itself an isometry) maps ABC to DCE, therefore these triangles are congruent. (The composite of two half-turns is always a translation.) Even though the Common Core is all about isometries, using this to answer the SBAC question is definitely not recommended.

Unfortunately, neither the girl nor the guy from the Pre-Calc class supplies the correct proof. The girl marks the congruence relations AD = BC and AB = DC on her paper, then proclaims "Yes correct." The guy doesn't even attempt this problem.

Both of these are Pre-Calc juniors. Neither has sat in a Geometry classroom in two years, since the end of their freshman year. So it's likely that both have forgotten how to write two-column proofs. These students need a refresher on Geometry proofs in order to be successful.

The marks on the girl's paper, by the way, suggests another valid proof based on SSS. If she had drawn in the congruence relations on the sides of parallelogram ACED, she would have had enough information to prove the triangles congruent:

Proof:
Statements                         Reasons
1. bla, bla, bla                    1. Given
2. ABCD is a pgram.         2. A rectangle is a parallelogram.
3. AB = DCAC = DE,       3. Opposite sides of a pgram are congruent.
    BC = AD, AD = CE
4. BC = CE                         4. Transitive Property of Congruence
5. Triangle ABC = DCE     5. SSS Congruence Property [steps 3,3,4]

This might be a better proof to show Algebra II or Pre-Calc juniors before the SBAC, since we don't have to remind them what Alternate Interior Angles are. On the other hand, my original proof is excellent for Geometry students before the final, since we want to remind them what Alternate Interior Angles are.

Question 28 of the SBAC Practice Exam is on nonlinear functions:

Choose the domain for which each function is defined.

[The four functions are f (x) = (x + 4)/xf (x) = x/(x + 4), f (x) = x(x + 4), 4/(x^2 + 8x + 16). The possible domains are all real numbers, x != 0, x != 4, x != -4. Here the "!=" means "does not equal" in ASCII and several computer languages.]

These functions are not linear. It will be hard-pressed to consider this an Algebra I question -- even though many texts (such as the Glencoe Algebra I text) contains a lesson on rational functions (Lesson 11-2), many teachers (including those in our district) skip Chapter 11 altogether. Thus even though students don't have to graph anything, this is mostly likely considered an Algebra II problem.

The first function has x in the denominator, so its domain is x != 0.
The second function has x + 4 in the denominator, so its domain is x != -4.
The third function has 1 in the denominator, so its domain is all real numbers.
The fourth function has (x + 4)^2 in the denominator, so its domain is x != -4.

Today is an activity day. The activity for Question 27 comes from Lesson 7-6 of the U of Chicago Geometry text, since this is the lesson on properties of a parallelogram and congruent triangles. The Exploration Question here is on congruent triangles in a regular pentagon. The activity for Question 28 comes from Lesson 13-4 of the U of Chicago Algebra I text, "Function Language," since this lesson discusses the domain and range of a function -- including rational functions like x/(x + 4). The Exploration Question here is about the range of the absolute value function. It's a great question for Common Core, since it involves translating the function up and down.

The guy from the Pre-Calc correctly answers all parts of this question. But the girl marks two boxes for the first part -- x != 0 and x != -4 -- and none for the second part. It could be that she just marks the answer for the second function in the first line instead. I'm not quite sure what would happen on the SBAC interface if a student tries to mark two answers in the same row -- it could be that each row is a radio button where only one box can be marked at a time. This would protect the girl from making this sort of mistake on the computer.

It's also interesting to see the two students' thought process for the fourth function. The guy factors the denominator as (x + 4)^2, then starts to mark x != 4 before crossing it out. On the other hand, the girl doesn't factor at all and just plugs x = -4 into the denominator instead. The guy's method is considered more sound, since the girl is relying on the multiple choice format. She would have had trouble with this question if she'd been asked to state the domain without multiple choice.


SBAC Practice Exam Question 27
Common Core Standard:
Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.

SBAC Practice Exam Question 28
Common Core Standard:
For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.

Commentary: The properties of parallelograms and congruent triangles are easily found in Chapter 7 of the U of Chicago Geometry text. Even though rational functions are usually saved for Algebra II, the domain and range of functions -- including rational functions as one of the examples -- appear in many Algebra I texts, including the U of Chicago text.



Thursday, May 23, 2019

SBAC Practice Test Questions 25-26 (Day 173)

In my old district -- the one whose calendar the blog is following -- today is the last day before the four-day Memorial Day Weekend. Last year, I explained why there was no school on Friday, May 4th of that year -- it was a "floating holiday." In years when Easter is early, the floating holiday in early May breaks up the long stretch between Easter and Memorial Day. But in years when Easter is late -- such as this year -- the floating holiday is placed right before Memorial Day to make it easier for holiday travelers to beat the traffic.

Thus there will be no post on the blog tomorrow. It's a shame since tomorrow in my new district -- which will still be open tomorrow -- I will sub in a middle school math class. It's the sort of day that I'd like to describe here on the blog, but unfortunately it lands on a non-blogging day just because the other district happens to be closed.

On the other hand, today I subbed in a seventh grade science class. And so I am doing "A Day in the Life" today, first because it's middle school where I can focus on my classroom management, and second because it's a science class which I can compare to the (lack of) science I taught back at the old charter school.

Before I start, I point out that this regular teacher will be out tomorrow as well -- but of course, this isn't a multi-day assignment because I'm already covering math (at a different school) tomorrow. Still, keep this in mind as you read today's "A Day in the Life."

8:00 -- Once again the regular teacher has morning yard duty, which I must complete.

8:15 -- The day begins with homeroom. This is the middle school where all periods rotate, but today the classes just happen to be in their natural order 1-2-3-4-5-6. And indeed homeroom is not the same as first period here -- generally homeroom matches the third period class.
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8:20 -- Homeroom leaves. First period is the regular teacher's conference period.

9:20 -- Second period arrives. All classes are seventh grade science today.

Since there is no science text, we expect there to be some assignment on Chromebooks -- and indeed, it's on endangered species. Students must choose an endangered species and answer the questions in Google Classroom on their chosen species.

The second assignment is a written worksheet titled, "Got Habitat?" Students answer questions on the habitats of cougars in the Santa Ana Mountains right here in Southern California.

Finally, the students have a crossword puzzle to complete. All the assignments are to prepare the students for tomorrow's test.

I decide to place the students complete all three assignments on my good list, while to avoid the bad list, the first two assignments must be finished. This is convenient because it's difficult for me to check the first assignment on the Chromebooks, but it's easy to check the second one on paper.

But no one finishes the third assignment, the crossword puzzle. Meanwhile, a whopping ten students fail to complete the second assignment, thus making a very long bad list. One of them is a boy who pretends that he can only speak Spanish when I ask him to stop playing Minecraft on a Chromebook.

10:10 -- Second period leaves for snack.

10:20 -- Third period arrives. There is a brand new student transferring into this class. It's difficult for him to work since he doesn't have a Google Classroom password. Three girls try to help him out.

Once again, no one can finish the crossword puzzle. It's tricky because the students can't even figure out the first clue, which is a five-letter word for a place where organisms live. It might be biome, but this isn't a word that they've learned in this current unit.

Thus I lower the bar for my good and bad lists. Now students who complete only two assignments are on the good list, while to avoid the bad list they must finish "1+" assignment -- that is, they must finish the first assignment online and have at least one valid answer on the second worksheet. And so I'm able to write some names on my good list for third period -- and I go back and retroactively add four names from second period to my good list as well.

I make it 1+ rather than just one to avoid the bad list because I want to see something written on the actual paper rather than rely on just an online assignment. In hindsight, I wonder whether I should have made it two to avoid the bad list and 2+ for the good list -- that is, at least something on the crossword puzzle must be filled in. But with eight students in this class failing to clear even the 1+ bar, there was no need to make it two full assignments. And this doesn't even include the three girls who claim that they're helping out the new student but don't even finish their own first assignment.

11:15 -- Third period leaves and fourth period arrives. And this class begins with yet another argument about the seating chart.

The regular teacher specifies that there is a seating chart to use. But as usual, students claim that the seats have been changed and the teacher never marked it on the seating chart. In those earlier classes, I take the liberty of creating a new seating chart based on where they actually are sitting. The idea is to leave it for tomorrow's sub so that no one will try to change seats during tomorrow's test. But then in each class after I create the chart, another student tries to switch seats -- and their justification is that the original seat changers have lied about the changes. In other words, why should they have to adhere to the chart if the others don't?

So when fourth period arrives, I seek to avoid students who lie about the seating chart. I announce that anyone not following the chart will receive an hour of detention. But then many students start to complain that they don't even remember where they sit because everyday, the regular teacher just says "Sit wherever you want!" and never punishes anyone for being in another seat.

At this point an eighth grader -- the TA for this class -- chimes in. He tells me an open secret -- the regular teacher plans on leaving this school at the end of the year. He probably has already made the decision months ago but is just waiting for the year to end. In many ways, he's already checked out and is no longer strongly interested in classroom management or student success. Indeed, the reason he's out today is a science meeting, but then he takes tomorrow off as well. (Once again, this isn't the other district where tomorrow's a holiday.) I don't want to speak ill of other teachers, but this is the sort of teacher the traditionalists often warn us about. (So tomorrow is spring Floyd Thursby day?)

Sometimes when I focus on improving my own classroom management, I tend to assume that all other teachers have perfect management skills, but this isn't always the case. It's difficult, if not impossible, to enforce a rule that the regular teacher himself doesn't enforce. He might have had a seating chart that he enforced at the start of the year before he made his decision to leave, but once again he doesn't enforce it now. The question I wonder is, how will the lack of a seating chart affect tomorrow's sub and test?

I think back to my own days at the old charter school -- where as you already well know, my management was less than perfect. Now suppose on the day of a test, I became ill, and so a sub had to give the test. How would my old students have behaved?

Actually, the seating chart was one of the rules I meticulously enforced that year. But of course, the students were very loud during the test. So my students could have claimed to that sub, "Our teacher, Mr. Walker, lets us talk during tests." And it would have been impossible for a sub to counter that claim enough for them to be quiet during the test.

But here's another fictional scenario regarding today's class. Suppose that new student in third period has accommodations that require him to sit near the front of the class -- and for him not to be told the reason why. (Once again this is a counterfactual -- I don't know of any accommodations needed for the new student in class today.) The best way to handle this is to hide his accommodation under the general rule "sit in your assigned seat" -- but this teacher doesn't enforce seat assignments. Thus we're in a bind, since of course the student would complain, "Why do I have to sit in the front when everyone else gets to sit anywhere?"

Now let's return to reality and what actually happens today. An argument is starting, and I've resolved to avoid arguments. So how can I get out of this one? I decide that since today isn't the actual test, it's not as important to enforce a seating chart as it will be for tomorrow's test. Then I leave a note for tomorrow's sub that the regular teacher doesn't enforce the chart, so it's up to that sub's discretion how to handle seating for the test.

After the argument ends, the class is a bit subdued. This actually ends up being the quietest class of the day, though I wish I could have achieved this without an argument. What doesn't drop is the number of students who fail to finish 1+ assignment -- another nine names go on the bad list.

12:05 -- Fourth period leaves for lunch.

12:50 -- Fifth period arrives. The class after lunch at this school is silent reading -- but since there are already Chromebooks out on the desks, some students start using them instead of reading. Someone (likely multiple kids) play YouTube videos one second at a time -- just to make noise during the silent reading and get everyone else to laugh.

Since so far I've written down too many names on the bad list, for this class I decide to return to my musical incentive. This time I choose Square One TV's "Mathematics of Love." I sing one verse at the start of class and tell them that there must be no more than one student on the bad list for me to sing the rest of the song. But it doesn't work -- another ten kids fail to complete 1+ assignment. On the other hand, it's the hardest-working class with about half the students finishing two assignments.

2:00 -- Fifth period leaves and sixth period arrives.

This time, many students don't want to work. By this period, I'm no longer enforcing a seating chart, but this means that the students take "sit wherever you want" up to the next level. Now many of them are no longer sitting in seats -- that is, there are 5-6 sitting down in a group that's designed for four, or they switch seats multiple times during the class. And these seat switchers don't do the work -- they either play games on the Chromebooks or just open up the first assignment and then talk instead of researching an animal.

For this class, only two students make the good list with two assignments, and only about a third of the class seems to be working at all.

At this point, I don't want to write two-thirds of the class on the bad list. So instead, I pass out the second worksheet and go over the answers on the board. Students must copy my answers onto their paper in order to avoid the bad list. Here are the three questions:

1. List several reasons why wildlife habitat is shrinking.
2. Why do cougars prefer riparian habitat?
3. What is the meaning of corridors used in this exploration?

Notice that this runs afoul of another resolution:

5. Engage the students in the learning process instead of lecturing excessively.

The situation today is similar to the one at the old charter school that led me to include this in the list of resolutions. In both cases, students are supposed to work independently, but most aren't. Unless I give them the answers to copy down, they're going to leave the paper blank. And while copying answers isn't the same as learning, at least having my answers on the paper will give them something to study for tomorrow's test, while a blank paper will give them nothing to study.

Notice that in all classes, the only assignment I actually check to determine the good and bad lists is the second assignment. In most cases, I'm using the honor system and assuming that if the students claim that they finished the first assignment online that they actually did. Yet most students in sixth period don't even pretend to be working on the first assignment online!

Well, you can't blame this class for acting as if school is already out After all, the regular teacher seems to be doing the same!

After we go over the three questions, less than one minute remains in class. The students ask me to sing "Mathematics of Love" -- and technically they did work, though not independently. There's only enough time to sing the first verse again -- which is good, since they haven't earned another verse.

2:55 -- Sixth period leaves. I put the Chromebooks away before leaving.

I feel sorry for tomorrow's sub. These are the classes that he or she must deal with and get them to be quiet for the test. What's worse is that due to the period rotation, sixth period will be the class that meets in the after-lunch silent reading position. After today's sixth period class, what are the odds that this class will be silently reading for a sub tomorrow?

Suppose I were returning to this class tomorrow. I ask myself, how would I handle it, knowing exactly what's coming?

This might be the time for a candy incentive. Normally I save the big candy incentives for holidays -- the ones associated with candy, such as Halloween or Christmas. There really isn't any Memorial Day candy (though some companies have started making Fourth of July candy and then selling it in time for Memorial Day). But there is precedent for me to bring candy to an especially difficult class. I once brought some to a class on the day between the regular teacher leaving and a long-term sub taking over. This is arguably a similar situation.

I would begin tomorrow's classes by handing out candy to all students on my good list. Depending on how much candy I purchase, I could even give two candies to those on the good list and one to those who avoid the bad list.

Then another candy would be given to each student who silently completes the test. Many might not like the idea of rewarding acceptable behavior (as opposed to outstanding behavior), but in a class where the regular teacher is leaving, there's really no other way to get the students to exhibit acceptable behavior.

The music incentive might fit here as well -- after everyone has finished the test and there's no problem with talking during the test, I sing the entire "Mathematics of Love" song.

OK, that ends what I would have tomorrow. The next question is, what should I have done today to get through today's classes better. What adjustments could I have made as the day progressed to get the students to do the work.

First of all, it would have helped if I knew the students' names -- but since none of them are following a seating chart, I don't know them. Notice that in periods 2-3, I have their names because I took the liberty of making a seating chart, but I stop after the fourth period argument and I find out that they're allowed to sit in any seat. But nothing is preventing me from creating a seating chart anyway. This time I wouldn't leave it for tomorrow's sub to enforce during the test -- it's just for my information to help me manage the class.

Also, I can change the order of the assignments. I've done so in the past when the second assignment requires extra explanation, but I fear that they won't reach it after doing the first assignment.

I can still have the students begin with the online assignment while I'm going around the room to create the seating chart. But then after I take attendance, I would immediately pass out the second worksheet and have them copy the answers from the board. If there's time left, they could return to the online assignment and then attempt the crossword.

The music incentive fits this plan better. One problem I have today with the song is that it requires too much from the students to obtain the incentive. Each student starts the class not working, and by I remind them about the incentive, class is half over and he or she feels there's not enough time to complete 1+ assignments. And even if a student thinks there is enough time for him or her to do it, the assumption that most of the others won't do it, so the incentive still won't be attained. Thus there's no point in that individual even trying.

One thing I like about the Quadratic Weasel song is that since it's so short, I can sing it multiple times during the class. I can go over each quadratic equation and sing it after every problem -- I just check all the papers to make sure all students have completed it. Thus today I could have sung a verse of the song after each of the three questions on the worksheet and checking all the papers.

Indeed, "Mathematics of Love" also fits this plan. Just as in the YouTube video, I open the class with the Roman numeral version of the song ("eye night, eye-eye hearts, III words"). Then after checking each question, I sing another verse of the song pronouncing the numbers correctly in English. Since there are three questions, we can consider the first "verse" starting with the first line ("one night, two hearts, three words"), the second "verse" (actually a bridge) with "1-2-3-forever," and the third verse with "7-8-9-tenderly."

Is it possible to establish a good and bad list under this plan? Well, I might be able to get some students to answer each question, so then I have at least three names on the good list. It might be possible to add names for finishing the online assignment -- but this time I'd have to check to make sure that the online assignment is actually complete. As we've seen before, there's no point in using the crossword as a criterion for the good list, since no one will finish it. As for the bad list, clearly those who don't write down answers the three questions belong there -- but they'd now have a bigger incentive to write, since they don't want to be the individual who blocks me from singing. The hope is that no one would land on the bad list under this plan.

I've written so much about classroom management that I've said little about the science. Anyway, assume that the teacher announced his intention to leave around the same time of year that I left the old charter (around late February or early March). Now compare how much science these seventh graders learned from August to February to how much my seventh graders from two years ago did. It is no contest -- my students learned much less, especially the seventh graders who learned hardly any.

Both this district and the old charter lack written science texts. At this school, I hear that even tomorrow's test is to be given online. I don't recall whether the Illinois State text required the science tests to be taken online. But once again, it shows that my failure to embrace technology and the online science curriculum is what cost me in the end.

This is what I wrote last year about today's lesson:

Question 25 of the SBAC Practice Exam is on inequalities:

A student earns $7.50 an hour at her part-time job. She wants to earn at least $200.

Enter an inequality that represents all of the hours (h) the student could work to meet her goal. Enter your response in the first response box.

Enter the least whole number of hours the student needs to work to earn at least $200. Enter your response in the second response box.

This is a first-semester inequality problem.

7.50h > 200
h > 26 2/3

So the inequality is 7.50h > 200 ("at least" = "greater than or equal to") and the smallest whole number value that satisfies it is 27 hours.

Both the girl and the guy from the Pre-Calc class fail to write the correct inequality. Both of them are confused by "at least" and write "7.50h < 200." Actually, the guy writes 200 > 7.50x, which is equivalent to the girl's inequality. The guy doesn't bother solving his inequality. The girl solves it by writing "at least 26.67" hours -- she missed the words whole number. Notice that no inequality needs to be written in the box -- all that's necessary is the single whole number 27.

Question 26 of the SBAC Practice Exam is on comparing statistics:

Michael took 12 tests in his math class. His lowest test score was 78. His highest test score was 98. On the 13th test, he earned a 64. Select whether the value of each statistic for test scores increased, decreased, or could not be determined when the last test score was added.

(The possible stats are standard deviation, median, and mean.)

Stats appears in the last chapter of Glencoe Algebra I, and so this is a second-semester question.

Anyway, the mean must decrease because Michael's last test score is lower than that of any previous test that he has taken. The standard deviation must increase when that low value is added. But we can't be sure about the median. The median of 12 values is the mean of the 6th and 7th value, but the median of 13 values is the 7th value (the old 6th value before the 64 happened). So median could decrease if the original 6th test is less than the 7th -- but the median could stay the same if the original 6th and 7th scores were equal. (It's impossible for the median to increase here!)

The girl from the Pre-Calc class gives the correct responses for the mean and median, but she incorrectly writes that the standard deviation decreased. Then again, most students are less familiar with standard deviation than with the measures of central tendency. But at least she tries -- the guy just leaves it blank.

Today is an activity day, since this is the last blog entry of the week. I decided to include the Exploration Question from Lesson 1-2 of the U of Chicago Algebra I text, since that is the lesson mentioned below. But since that lesson doesn't mention standard deviation, I place a short activity on standard deviation on the other side. The students must at least know how to find standard deviation on a calculator to complete it.

SBAC Practice Exam Question 25
Common Core Standard:
Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

SBAC Practice Exam Question 26
Common Core Standard:
Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.

Commentary: The inequality I wrote for h can be solved as early as Lesson 4-6 of the U of Chicago Algebra I text. Meanwhile, stats isn't covered in the text at all, except for Lesson 1-2 where mean and median appear, but not standard deviation.

My next post will be on Tuesday. I hope you enjoy your Memorial Day weekend, since I certainly will -- but not until after tomorrow's math class, of course.