Tuesday, January 15, 2019

Famous Movie Math Quotes #5

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(C)Copyright 2018, C. Burke.

"We'll always have Pairs" is #43.

I pared down the dialogue a bit. It wouldn't all fit, and it wouldn't all be funny or particularly mathy.

For the newcomers, the number 5 in the title is a reference to the AFI's Top 100 Movie Quotes list from ten years ago or so. I've done a couple of these over the years, but not in any particular order. Will I do more of these? Maybe not today, maybe not tomorrow, but soon ... or not so soon.




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Friday, January 11, 2019

Clip - Art

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(C)Copyright 2018, C. Burke.

That Art! He is the Clipper to the Stars!

Ten years in the making, at least. And I originally wanted to put some familiar head shots in there, like you see on barber walls.

I've had the joke, but not the means to do it. Then, on a whim, I checked the web for 8-bit creators, and found a couple. I created a witch in no time. And then I nearly crashed my iPad when I tried using some of the site's functions to either save or email the picture. Seriously, if was a half-hour or more before my iPad was responding, and I was afraid I would have to do another factory reset.

Luckily, the second site was a little better and PC friendly. I recreated my witch, print-screened and saved it that way. Then I re-colored it to match my original witch, which I liked better. (Which witch? The first one.)

I made the next three in a couple of days, and passed them by the expert: my son. I thought they came out good, but didn't know what was sticking on the side of the second guy's head. I said I was going for "Elf". Oh, well. I changed the coloring a little.

Next problem was that the joke was ALWAYS "Four SHAVES, Four Haircuts". Only one guy had a beard -- and one of the others was female.

First try was to create the pirate, who I thought was "okay" but I wasn't happy with his eyepatch or his hat. Keep in mind, he came before either of the two pirate jokes that I already posted this week. He could've tied them all together very nicely. (He might've been part of the reason I came up with the other two jokes, and well as the play on "Barber".)

But I liked the witch, and I thought that a wizard wouldn't be a difficult transition. So I went that route instead.

For those who are interested, the original artwork is posted below. The knight always had a beard, even before I "remembered" the joke. Witch, elf, barbarian (part satyr?), knight, pirate, wizard.

I hope you like them. I could try to recreate them sometime, but it won't be easy though.




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Wednesday, January 09, 2019

Barbary Pirates, Part II

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(C)Copyright 2018, C. Burke.

Many a battle were close shaves!

I apologize for nothing.




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Tuesday, January 08, 2019

August 2018 Algebra I Regents, Part II

The following are some of the multiple questions from the August 2018 New York State Common Core Algebra I Regents exam.

August 2018 Algebra I, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.


25. Explain how to determine the zeros of f(x) = (x + 3)(x -1)(x - 8).
State the zeros of the function.

Answer:
To find the zeroes of a function in factored form, set each of the factors equal to 0 and solve for x.

The zeroes of this function are:


x + 3 = 0 OR x - 1 = 0 OR x - 8 = 0
x = -3 OR x = 1 OR x = 8



26. Four relations are shown below.


State which relation(s) are functions.
Explain why the other relation(s) are not functions.

Answer:
III and IV are functions.
I is not a function because it fails the vertical line test. When x = 6, there are two values of y, 4 and 6. This is not allowed.
II is not a function because the x values 1 and 2 are repeated with different y values.

Added information: In III, the x values in the table do not repeat. In IV, the quadratic equation given is a parabola, which is a function. There is only one possible outcome when you square a number.


27. The table below represents the height of a bird above the ground during flight, with P(t) representing height in feet and t representing time in seconds.

Calculate the average rate of change from 3 to 9 seconds, in feet per second.

Answer:
P(3) = 6.26 and P(9) = 3.41.
The average rate of change is (3.41 - 6.26) / (9 - 3) = -2.85 / 6 = -0.475 feet per second.


28. Is the solution to the quadratic equation written below rational or irrational? Justify your answer.

0 = 2x2 + 3x - 10

Answer:
Check the discriminant: b2 - 4ac
32 - 4(2)(-10) = 9 - (-80) = 89
The square root of 89 is an irrational number, so the solutions to the equation are irrational.


29. The formula for converting degrees Fahrenheit (F) to degrees Kelvin (K) is:

K = (5/9) (F + 459.67)
Solve for F, in terms of K.

Answer:

K = (5/9) (F + 459.67)
(9/5) K = F + 459.67
(9/5) K - 459.67 = F



30. Solve the following equation by completing the square:

x2 + 4x = 2

Answer:
Take half of 4 (which is 2) and square it (which is 4). Add that to both sides.

x2 + 4x = 2
x2 + 4x + 4 = 2 + 4
Now factor
(x + 2)2 = 6
x + 2 = + SQRT (6)
x = -2 + SQRT (6)

Since the wanted an exact value, leave the answer in fraction form because it's a repeating decimal. DO NOT ROUND.


31. The students in Mrs. Lankford’s 4th and 6th period Algebra classes took the same test. The results of the scores are shown in the following table:


Based on these data, which class has the largest spread of test scores? Explain how you arrived at your answer.

Answer:
The 4th Period has the larger spread.
The sigma value (the second column), which is used to find variance and standard deviation, is greater for 4th period, so the spread will be greater.
Also, if you calculate the Interquartile Range, Q3 - Q1, you will get the following results:
4th Period: 87.5 - 69 = 18.5
6th Period: 88 - 71.5 = 16.5
4th Period has the greater IQR so it has the greater spread of the data.

Personal note: I'm not fond of this question. A little to vague. However, they chose data that would give the same answer whichever method was chosen. I'm not sure if calculating only the range would have been good enough, but if you search online, you will see that the range is listed as one measure of the spread of data. (Certainly, the others are better measures, but it doesn't mean the other method isn't one.)


32. Write the first five terms of the recursive sequence defined below.

a1 = 0
an = 2(an - 1)2 - 1, for n > 1

Answer:
a1 = 0
a2 = 2(0)2 - 1 = 0 - 1 = -1
a3 = 2(-1)2 - 1 = 2(1) - 1 = 1
a4 = 2(1)2 - 1 = 2(1) - 1 = 1
a5 = 2(1)2 - 1 = 2(1) - 1 = 1

0, -1, 1, 1, 1

Note: every term after the 3rd will be 1.

End of Part II

How did you do?

Questions, comments and corrections welcome.

Monday, January 07, 2019

(x, why?) Mini: Barbary Pirates, Part I

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(C)Copyright 2018, C. Burke.

Ken you defeat the Barbarie Dream boat? You can really test you Mettle.

Funny, I work on dialogue and/or narration for a while, as well as considering props ... and then just kept it simple.




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Thursday, January 03, 2019

Happy Thirdsday

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(C)Copyright 2018, C. Burke.

Enjoy it before it gets over-commercialized! Only happens on Thursday, on average, once every seven years!

We don't get another one until 2030!




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Tuesday, January 01, 2019

Happy New Year 2019!

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(C)Copyright 2018, C. Burke.

Happy New Year! I'm getting tired of these.

Seriously, if I kept a regular schedule and produced close to 150 comics per year, then "filler" comics like this one would be okay. But I've only done 1400 or so in 11 years, which is a smaller average.

I don't want to be a holiday comic any more than I want to be an "in memorium" comic. Basically, I need a fresh take on New Years Eve and New Years Day. The "Eve" is the party day, while the "Day" is the number change. I've done binary and other weird combinations already.

And I'm less impressed about finding some mathematical equation that equals 2019 (or whichever year) because in the waning days of December, there will be tweets galore on social media with so many sums of squares and such. (And quite a few will be just expanded binary operations.)

In summary, if something brilliant doesn't occur to me -- and frankly, I don't think I'll ever top 2013 or the original bad pun for 2008 -- next year will likely be a random character stating "Happy New Year. This is 2020." Okay, I guess I need a Hugh Downs and Barbara Walters by then.




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Monday, December 31, 2018

New Years Eve Plans!

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(C)Copyright 2018, C. Burke.

Now I know why I like math puns so much. When people talk, it gets so wordy!

And there's a lot of standing around so that they get a chance to say this stuff.

And I deleted and rewrote a lot of dialogue.

Which is why this didn't actually appear on the last day before Christmas break!

I considered labeling this one "School Life", but I wanted to keep that for the students. And I've toyed with "After Hours" for the lives of the characters, but this is obviously school time. When else would Mike and Judy be around each other? Such is the life of a comic writer.

Oh, and Happy New Year! May all your plans be successful!




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Friday, December 28, 2018

(x, why?) Mini: Shapes Are Cool!

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(C)Copyright 2018, C. Burke.

Their shapes are compatible though trapezoids are less elegant. Trapezoids have no concept of elegance.




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Thursday, December 27, 2018

Algebra 2 Problems of the Day

(not quite) Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part IV

The Question in Part IV is worth up to 6 credits. Partial credit is possible.


37. A radioactive substance has a mass of 140 g at 3 p.m. and 100 g at 8 p.m. Write an equation in the form A = A0(1/2)(t/h) that models this situation, where h is the constant representing the number of hours in the half-life, A0 is the initial mass, and A is the mass t hours after 3 p.m.

Using this equation, solve for h, to the nearest ten thousandth.

Determine when the mass of the radioactive substance will be 40 g. Round your answer to the nearest tenth of an hour.

Answer:
We are given A = 100, A0 = 140, and t = (8 - 3) = 5. We want h.
A = A0(1/2)(t/h)
100 = 140 (1/2)(5/h)
100/140 = (1/2)(5/h)
5 / 7 = (1/2)(5/h)
(log 5/7) / (log 1/2) = (5/h) log (1/2) / log (1/2)
(log 5/7) / (log 1/2) = 5 / h
(log 5/7) * h = (log 1/2) * 5
h = (log 1/2) / (log 5/7) * 5
h = 10.3002, to the nearest ten-thousandth.

Second part:
We are given A = 40, A0 = 140. We now have h = 10.3002. We want t.
A = A0(1/2)(t/h)
40 = 140 (1/2)(t/10.3002)
40 / 140 = (1/2)(t/10.3002)
2 / 7 = (1/2)(t/10.3002)
(log 2 / 7) / (log 1/2) = t / 10.3002
t = 10.3002 * (log 2/7) / (log 1/2)
t = 18.61... = 18.6 hours to the nearest tenth of an hour.


Logs were never my favorite subject in school, even if the rules are pretty straightforward. It's just one more inverse operation.

Next up: January 2017.



Comments and questions welcome.

More Algebra 2 problems.

Wednesday, December 26, 2018

Can Hardly Wait!

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(C)Copyright 2018, C. Burke.

Plot twist: Mom works outside of the house, too, but she took off this week to spend time with the kids!




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Tuesday, December 25, 2018

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part III

All Questions in Part III are worth up to 4 credits. Partial credit is possible.


35. Graph y = log2(x + 3) - 5 on the set of axes below. Use an appropriate scale to include both intercepts.

Describe the behavior of the given function as x approaches -3 and as x approaches positive infinity.

Answer:
If you graph y = log2(x + 3) - 5, you will see that it is not defined for x %lt 3.
The y-intercept is at y = log2(0 + 3) - 5 = -3.42.
Solving 0 = log2(x + 3) - 5 gives you the x-intercept at 29.
You can use the calculator, or work it out:

0 = log2(x + 3) - 5
5 = log2(x + 3)
x + 3 = 25
x + 3 = 32
x = 29

The best scale to use for the x-axis is 3. Use a scale of 1 for the y-axis.
Check the image below. The table of values is provided for you. It wasn't necessary for the exam, but it would be a good idea to label the intercepts.

Second part: As x approaches -3, the function goes to negative infinity. As x approaches infinity, the function approaches infinity.



36. Charlie’s Automotive Dealership is considering implementing a new check-in procedure for customers who are bringing their vehicles for routine maintenance. The dealership will launch the procedure if 50% or more of the customers give the new procedure a favorable rating when compared to the current procedure. The dealership devises a simulation based on the minimal requirement that 50% of the customers prefer the new procedure. Each dot on the graph below represents the proportion of the customers who preferred the new check-in procedure, each of sample size 40, simulated 100 times.


Assume the set of data is approximately normal and the dealership wants to be 95% confident of its results. Determine an interval containing the plausible sample values for which the dealership will launch the new procedure. Round your answer to the nearest hundredth.

Forty customers are selected randomly to undergo the new check-in procedure and the proportion of customers who prefer the new procedure is 32.5%. The dealership decides not to implement the new check-in procedure based on the results of the study. Use statistical evidence to explain this decision.

Answer:
The mean is given as 0.506, and one standard deviation is 0.078. To be 95% confident in the result requires two standard deviations, or 2 * 0.078 above or below the mean. So the interval would be:

0.506 - 2 * 0.078 < 0.506 < 0.506 + 2 * 0.078
0.506 - 0.156 < 0.506 < 0.506 + 2 * 0.156
0.354 < 0.506 < 0.656
0.35 < 0.506 < 0.66
Be sure to round to the nearest hundredth.

In the second part, since 32.5% is below 35.4%, the amount is outside of the 95% confidence interval.



Comments and questions welcome.

More Algebra 2 problems.