Posts tagged with: OT Test 1 Section 4

Test 1 Calculator OK #16

The graph here is definitely a linear relationship between h and C, so we’re going to want to figure out the slope and the C-intercept. Then we’ll use the standard y=mx+b slope-intercept formula, only for this problem it’ll be C=mh+b.

Slope is rise over run, or the change in the vertical variable divided by the change in the horizontal variable. Let’s pick two points from the line and plug them into the slope formula. How about (1, 8) and (3, 14)?

    \begin{align*}\text{slope}=\dfrac{C_2-C_1}{h_2-h_1}=\dfrac{14-8}{3-1}=3\end{align*}

Once we have the slope, we’re actually done because whoever wrote this question wasn’t very tricky with the answer choices. The C-intercept of this graph is obviously not zero, and only choice C has the slope we just calculated plus a non-zero C-intercept. Sure enough, a quick glance at the graph confirms that the C-intercept is 5, meaning the equation should be C=3h+5.

Whats the best and fastest way of doing Practise test 1, section 4, question 23?

Would it be calculating the ratio of all the options and then comparing it to the Human Resources ratio?

Thanks!

Yeah, that’s the best and fastest way, although you can save yourself some calculator keystrokes by rounding. This isn’t just a time saver: the more keystrokes you make, the more likely you are to make an error! Rounding to the nearest million (it’s already in thousands) will work just fine. So, my calculations would look like this:

Human resources 2007 to 2010: \dfrac{4051}{5921}=0.68

Agriculture/natural resources 2007 to 2010: \dfrac{374}{488}=0.77

Education 2007 to 2010: \dfrac{2165}{3008}=0.72

Highways and transportation 2007 to 2010: \dfrac{1468}{1774}=0.83

Public safety 2007 to 2010: \dfrac{263}{464}=0.57

There you have it: the Human resources and Education budgets have the closest ratios.

Hi Mike. Can you please explain Practise test 1, section 4, question 14? I figured out that the range was the largest change (but only by physically calculating the mean and median etc.)

Is there a better way to do this?

Also, I’d just like to say THANK YOU THANK YOU THANK YOU for this book it really is amazing!!

Well, you have to consider every answer choice, but there are a few mental shortcuts that I think are helpful.

The main thing to recognize right away is that the data is in order, so calculating the median is very easy. The middle value is 12, and if you remove that outlier 24, the median will STILL be 12. (It’s useful to remember that, in general, outliers have little to no effect on the median—that’s why you usually hear about median household income instead of mean household income, for example.) So I can cross of median without calculating it, and then I can cross off choice D as well.

The data being neatly sorted also helps one to think about the mean. Notice how nicely clustered the data is: how the 13s, 14s and 15s pretty nicely balance out the 11s, 10s, and 9s. That tells you that the mean might not be exactly the same as the median, but it should still be very close to 12. So without adding up all the values, I know the average of 20 of the values should be about 12, which means their sum should be about 240 (average times # of numbers equals sum). Adding the erroneous 24 value in there won’t change the average much: \dfrac{\approx 240+24}{21}\approx 12.57. (Note that while this estimate is not completely accurate, it doesn’t matter because you’re just looking for the biggest change, not the exact change.)

The range, on the other hand, will change a lot! The range with the 24 measure in there is 24-8 = 16. Take that 24 out and the range becomes 16 - 8 = 8. That’s by far the biggest change.

PS: Glad you like the Math Guide! 🙂

Test 1 #35 (calculator section)

You can find the formula for the volume of a right circular cylinder at the beginning of the section; it’s V=\pi r^2h. Since the question tells you the volume (72π cubic yards) and the height (8 yards), you can solve for the radius:

    \begin{align*}72\pi&=\pi r^2(8)\\9&=r^2\\3&=r\end{align*}

Now, be careful! The question asks for the diameter, not the radius. Even though they underline the word diameter, I bet a lot of people still miss this question by putting the radius.

If the radius is 3, then the diameter is 6.

Hello, can you please explain number 27 section 4 test 1?

Yeah, this is a really tricky one. The key to getting it right is recognizing just how much of the field the students are NOT covering! In a 10 ft by 10 ft field, there will be 100 1 ft by 1 ft squares. Look, below is a 10 by 10 grid with X’s marking places where the 10 students in this question might be.

The data in the table tells us that, on average, there are about 150 worms in each of the small boxes. Since there are 100 small boxes in the big field, we need to multiply 150 by 100 to get our answer for the approximate number of earthworms in the whole field.

150(100) = 15,000