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<p>Thank you!</p>

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<p>Game Attendance
1 456
2 508
3 399
4 550
5 n
The table above shows the attendance at the home games of the Central High School football team. If the median attendance for the five games was 456, and no two games had the same attendance, what is the greatest possible value for n?</p>

<p>the choices are (A) 400
(B) 455
(c)457
(D)478
(E)549
I understand why thing answer is B and I get why none of the other answers make sense but why can’t it be A?</p>

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<p>Putting the attendances in numerical order:
399,456,508,558 For the median to be 456, n has to be less than 456 ( and any number less than 456 will do). The question asks for the largest possible value of n. The largest integer (can’t have fractional people at a game) less than 456 is 455, which is B.</p>

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<p>@sararabinovich, questions’s asking for the largest possible n. B) 455 definitely works, so A) is incorrect.</p>

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<p>Oh! I didn’t realize it said largest. Haha I think the heat is getting to my head.</p>

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<p>Hiii!can you help me with this one please</p>

<p><a href=“http://i1069.■■■■■■■■■■■■■■■/albums/u477/natyyyyym/73ccbba0.jpg[/url]”>http://i1069.■■■■■■■■■■■■■■■/albums/u477/natyyyyym/73ccbba0.jpg&lt;/a&gt;&lt;/p&gt;

127

<ol>
<li>The average rate is d(t)/t, or a + (1/2)bt. Right away, this eliminates choices B, D, and E because they are not in the form (constant) + (constant)t. Also, it cannot be A, otherwise d(t) = 0 + t^2, and 0 is not a positive constant. Therefore, process of elimination, D.</li>
</ol>

<p>You can also show that D is right, since d(t) will equal t + t^2. It works for t=0, t=1, t=2, etc.</p>

128

<p>Marbles are to be removed from a jar that contains 12 red marbles and 12 black marbles. What is the least number of marbles that could be removed so that the ratio of red marbles to black marbles left in the jar will be 4 to 3? </p>

<p>The answer is supposed to be 3…help?</p>

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<p>Remove 3 black marbles and the ratio will be 12:9 which reduces to 4:3. Were you thinking that the marbles would be moved to some other jar?</p>

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<p>sorry, heres another: </p>

<p>The interior dimensions of a rectangular fish tank are 4 feet long, 3 feet wide, and 2 feet high. The water level in the tank is 1 foot high. All of the water in this tank is poured into an empty second tank. If the interior dimensions of the second tank are 3 feet long, 2 feet wide, and 4 feet high, what is the height of the water in the second tank?</p>

<p>A) 0.5 ft
B) 1 ft
C) 1.5 ft
D) 2 ft
E) 4 ft </p>

<p>Answer: E</p>

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<p>@6thstation it’s D. One way to think of it is, the volumes of both tanks are equal (24 ft^3). The first tank is half full, so the second tank must also be half full upon dispensing the water into that tank. Half of 4 is 2, so the height of the water in the second tank is D) 2 ft.</p>

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<p>@rspence: that’s a very nice explanation! Noticing that the volumes of the tanks are the same definitely leads to an easier solution…</p>

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<p>@pckeller yeah it definitely helps to notice that the volumes are the same, but 6thstation claimed it was E. Is that his answer, or some error from a review book?</p>

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<p>I apologize; the answer was D.
Thank you!</p>

135

<p>In the equation, k and m are constants. If the equation is true for all values of x, what is the value of m?</p>

<p>(x-8)(x-k)= x^2-5kx+m</p>

<p>A is 8, B is 16, C is 24, D is 32, E is 40</p>

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<p>(x-8)(x-k) = x² - 5kx + m
Multiplying out the factors on the left side
x² - 8x - kx + 8k = x² - 5kx + m
Now you can create 2 equations:
-8x - kx = -5kx and m = 8k
First solve for k.
-8x - kx = -5kx
-8x = -4kx (Add kx to both sides)
kx = 2x (Divide by -4 on both sides)
k = 2
Then use m = 8k
m = 8(2)
m = 16</p>

<p>checking your answer if you have time:
(x-8)(x-2) = x² - 10x + 16
x² -8x - 2x + 16 =
x² - 10x + 16 = x² - 10x + 16</p>

<p>:)</p>

137

<p>Um if anyone has the Blue book 2nd edition I need help with problem # 8 on page 374 thank you</p>

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<p>It says that line l is translated 2 units to the left, which means that the given point (-5,0) will be turned into (-7,0). Then you can plug a point into the equation that they give you.</p>

<p>y = .8x + k
0 = (4/5)(-7) + k
0 =(-28/5) + k
k = 28/5 or k = 5.6</p>

<p>Checking: Graph the equation y = .8x + 5.6 and check that the points (-7,0) and (-2,4) are on it.</p>

139

<p>If a triangle has exactly one of its vertices on a circle, which of the following CANNOT be the number of points that the triangle and the circle have in common? </p>

<p>A. Two
B. Three
C. Four
D. Five
E Six</p>

<p>The correct answer is six, and while I understand that it’s correct, I need evidence for why A. “Two” is incorrect? How can a triangle with one vertex on a circle possibly have two points in common with the circle?</p>

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<p>@yellowcat429, yeah, there is a way to get A) Two. Suppose that in triangle ABC, A is on the circle. AB could be tangent to the circle, AC could intersect the circle, and BC might not intersect the circle at all. Hence it’s possible to get exactly two intersection points.</p>