<p>1.)The image shows a bar graph titled: MILEAGE TEST RESULTS
Y axis: Number of Cars
X axis: Gas Mileage (miles per gallon)
Here is a data table of the values since I can't post the graph.
X Y
25 40
26 50
27 40
28 30
29 10
What is the median gas mileage for these cars, in miles per gallon?
A 26
B 26.5
C 26.75
D 27
E 27.5</p>

<p>In total there are 170 cars, so just start crossing off the lowest and highest MPGs. The 40 25s will knock out the 30 28s and 10 29s, and then it’s just finding the median between 50 26s and 40 27s, so the answer is then obviously 26.</p>

<p>A little less confusing method, but billabongboy is using a correct method to solving this problem. Nicely done and it’s very efficient.</p>

<p>1) How many cars are there total?
Answer: 40 + 50 + 40 + 30 + 10 = 170</p>

<p>Median is a type of average meaning the center of the graph. Two get the median you can keep going inwards until you reach the center, but an easier method is taking the total and dividing by 2. </p>

<p>2) 170/2 = 85
To reach 85 count inwards until you reach 85. 40 + 50 = 90. In this graph (I’m assuming bar graph), 85 is also equal to [41,90].</p>

<p>3) What is the gas mileage for the range [41,90]?
Answer: 26 mpg</p>

<p>I thought that Median is simply the middle of some set of numbers when you put them in order. like 1,2,3,4,5 then 3 would be the median because it’s in the middle. But why did you have to divide it by 2? Why would you need the total anyway? And what do you mean by “The 40 25s will knock out the 30 28s and 10 29s”?? I don’t understand what you guys did, so can you please explain a bit more to me.</p>

<p>I’m sorry if I’m being dumb, but I really really don’t understand this.</p>

<p>^Your definition is correct. It is a little trickier when you have an even number of items, when you have to take the average of the two numbers in the middle. For example, the median of 1, 5, 3, and 2 is 2.5, since the sorted list is 1, 2, 3, 5 and (2+3)/2 = 2.5.</p>

<p>To get the median gas mileage, you have to imagine building a list of all the gas mileages in increasing order. The list would look like: 25, 25, 25, … , 25, 26, 26, 26, …, 26, 27, 27, … and so on. Which is the middle number in this list? There are 170 total cars, which you can imagine as two groups: 1-85 and 86-170. So, we need the average of the 85th and the 86th cars’ gas mileage. This is just (26+26)/2 = 26. Hope that helps.</p>

<p>In general, for a list of N things: if N is odd, the middle element is (N/2) rounded up; if N is even, the middle elements are (N/2) and (N/2)+1.</p>

<p>I’ve got some more math questions that I can’t figure out how to do.</p>

<ol>
<li><p>The least and greatest numbers in a list of 7 real
numbers are 2 and 20, respectively. The median of the
list is 6, and the number 3 occurs most often in the list.
Which of the following could be the average
(arithmetic mean) of the numbers in the list?
I. 7
II. 8.5
III. 10
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
(E) I, II, and III</p></li>
<li><p>Each of the following inequalities is true for some
values of x EXCEPT</p></li>
</ol>

<p>(A) x < x^2 x^3
(B) x < x^3 < x^2
(C) x^2 < x^3 < x
(D) x^3 < x < x^2
(E) x^3 < x^2 < x</p>

<p>Why can’t E be a correct answer too? if you put in negative number for x, it isn’t true because (-x)^2 is not smaller than x, though the first part is correct…please help! </p>

<ol>
<li>
Family number of consecutive nights
Jackson 10
Callan 5
Epstein 8
Liu 6
Benton 8</li>
</ol>

<p>The table above shows the number of consecutive
nights that each of five families stayed at a certain
hotel during a 14-night period. If the Liu familys stay
did not overlap with the Benton familys stay, which of
the 14 nights could be a night on which only one of the
five families stayed at the hotel?
(A) The 3rd
(B) The 5th
(C) The 6th
(D) The 8th
(E) The 10th</p>

<p>Does question 1 have to be an integer in the list?</p>

<p>2,3,3,6,n,(n+1),20.</p>

<p>2+3+3+6+(n)+(n+1)+20/7, where n is 6<x<20.
You must have different numbers because 3 appears at most 2, which would mean the unknown numbers are forced to be n and n+1. </p>

<p>Take the average of it.</p>

<p>35+2n/7 = 5 + 2n/7 or (34+(N)+(N+1))/7, which are both the same thing. </p>

<p>5+2n/7 is easier to work with. I simply plugged in.</p>

<p>When n = 7, 5+2 = 7. When n=12.25 would yield an 8.5. And when n =17.5, you get 10.</p>

<p>It seems as though they’re only integers. I would say A. If not, then E. Not sure if any of this is correct, just my general thinking. Let me know what the correct answer is. Thanks.</p>

<ol>
<li>You can take one term and pretend it’s being multiplied by all the other numbers. Do this in each one and you will figure out that only 9 terms are divisble by 5. It doesn’t matter which order you do it, 5x10 and 10x5 counts for only 1 possibility. </li>
</ol>

<p>I got: 4 x 10, 5 x 10, 6 x 10, 5 x 11, 6 x 10. 5/9. I think the person above me may have miscounted or I am wrong somehow.</p>

<p>Put the numbers in order: 2,3,3,6,?,?,20.
Add the numbers, excluding the question marks. You get 34.
If you check the answers, you can multiply each of them by 7 because that’s the way you get the average/arithmetic mean. </p>

<p>For number 1, it would be 49. Number 2, 59.5. Number 3, 70. If it’s restricted to only integers, then only I and III. If not, all would.</p>

<p>The flaw in my little formula was that I assumed it would be an integer n+1 greater than the number before it but that’s not actually necessarily so. You could fill random numbers in there from 7 to 19 for each.</p>

<p>5) Find circumference of both semi circles
Let’s say the radius of the smaller circle is 4
and the radius of the bigger circle is 8
both radius add to 12 which is line RS</p>

<p>Circumference = 2 * radius * pi
since it’s a semi circle, you want half the circumference.</p>

<p>Small Circle :: 1/2 * 2 * 4 * pi = 4
Bigger Circle :: 1/2 * 2 * 8 * pi = 8</p>

<p>add both half circles’ cirumferences,
the answer is (C) 12 pi</p>

<p>I don’t think that works if you use radius=2 for the small one and 10 for the big one…I got 6pi for that…but shouldn’t all radii that add up to 12 work?</p>