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Michael C.
Math, Science, Programming, and Test Prep Expert
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SAT
TutorMe
Question:

If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m? A) m+6 B) m+7 C) 2m+14 D) 3m+21

Michael C.

First, we need to identify what this questions is asking. Most often, it's the last thing written in any SAT math problem. In this case, the question is "what is the average of x, y, and z in terms of m?" "In terms of m" means that the multiple choice options listed will all be expressions that contain the variable m. The test-makers love to use the phrase "in terms of..." at the end of math questions, since it often confuses students and makes them waste time. My advice? Ignore it. The choices given already tell us m needs to be there. The real question is "what is the average of x, y, and z?" As long as we remember how to average numbers, this is a pretty straightforward question. So what are x, y, and z? They tell us in the beginning of the problem! x is the average of m and 9, so all we need to do is add m and 9, and then divide by 2. We don't know what m is, but that's okay. We just need an expression. We do the same thing for y and z, and then we have this. $$x=\frac{m+9}{2}$$, $$y=\frac{2m+15}{2}$$, $$z=\frac{3m+18}{2}$$ Now, to average x, y, and z together, we just need to add them together... $$x+y+z=\frac{m+9}{2}+\frac{2m+15}{2}+\frac{3m+18}{2}=\frac{6m+42}{2}=3m+21$$ Then, divide by 3. $$\frac{x+y+z}{3}=\frac{3m+21}{3}=m+7$$ And that's it! The answer is B. NOTE: You might notice that we found 3m+21 halfway through solving the problem, and that expression is listed as choice D. With algebra problems, the test-makers will often include expressions that we find along the way as trap answers. Be sure you know what you are doing and when you have done everything you need to do to get the correct answer.

Trigonometry
TutorMe
Question:

What is the smallest positive value for x for which y = sin 5x is a maximum?

Michael C.

The sine function generally outputs a value between -1 and 1, inclusive. So if $$y = sin (x)$$, the maximum value we expect to find will be 1. By using the unit circle, we can easily recognize that $$x$$ must equal $$\frac{\pi}{2}$$ in this case. However, since $$y = sin (5x)$$ rather than $$sin (x)$$, we can find our solution by setting $$5x=\frac{\pi}{2}$$ and solving for $$x$$. Answer: $$x=\frac{\pi}{10}$$

Statistics
TutorMe
Question:

We are conducting a survey to determine the proportion of US Households that are "cable-cutters" (have Wi-Fi, but not cable television). We want like to create a 98% confidence interval with the margin of error of at most 4%. Assuming our population is normally-distributed, what is the minimum sample size we need to ensure our margin of error meets this requirement?

Michael C.

This is a question polling companies ask all the time. Surveying is expensive, and there is a constant tug-of-war between clients wanting both the most accurate survey results possible (which requires a large sample size) and keeping the cost of conducting the survey low (smaller sample = cheaper). Fortunately, it's fairly simple to calculate the minimum sample size for a normally-distributed population. Since we are looking for a proportion of US Households (rather than measuring a mean value), we use the following formula to calculate our minimum sample size: $$n=z^2p(1-p)/E^2$$ where p is the assumed proportion of cable-cutters (since this isn't given, we use .5 by default), E is our margin of error (we use .04, the decimal form of 4%), and z is the z-score for the 98% confidence level (we can find this on the z-table: 2.33). $$n=2.33^2*.5*(1-.5)/.04^2=848.2656=849$$ We must sample a minimum of 849 US Households in order to achieve the desired level of accuracy. NOTE: Notice that we rounded our final answer up rather than down. If we round down, our margin of error will exceed 4% and our survey will be less accurate than desired. Always round up when calculating minimum sample size.

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