Can someone help me with these math questions?

Test Preparation SAT Preparation
1

<p>they’re from some random sat test i got a hold of, so it’s not in the blue book.</p>

<li></li>
</ol>

<p>|X|-|-|X|</p>

<p>The figure above represents four offices that will be assigned randomly to four employees, one employee per office. If Karen and Tina are two of the four employees, what is the probability that each will be assigned an office indicated with an X?</p>

<p>A) 1/16
B) 1/12
C) 1/6
D) 1/4
E) 1/2</p>

<p>The answer is C, but I have no idea why</p>

<li></li>
</ol>

<p>2x - 5y = 8
4x + ky = 17</p>

<p>For which of the following values of k will the system of equations above have NO value?</p>

<p>A) -10
B) -5
C) 0
D) 5
E) 10</p>

<p>Answer:(A)</p>

<p>Why?</p>

2

<p>15.
When you double the first equation, you get 4x-10y=16. If you notice, the second equation also has a 4x, but the constant term is 17. To make this have no solution, just set the two equations "equal" to each other, because how can 16=17? It can't. Thus, giving no solution to the system of equations</p>

<p>I'll get back to you on the first one</p>

3

<p>20
You could list all the possible combinations which won't take too long.
Let's let Karen and Tina be A and D
ABCD BACD CABD DABC
ABDC BADC CADB DACB
ACDB BCDA CBDA DBCA
ACBD BCAD CBAD DBAC
ADCB BDAC CDAB DCAB
ADBC BDCA CDBA DCBA</p>

<p>Now, out of all those possiblities, how many have and A and D on both ends?
4. How many possibilities are there? 24. So 4/24 = 1/6</p>

4

<p>20) <a href="http://talk.collegeconfidential.com/archive/index.php/t-197040.html%5B/url%5D"&gt;http://talk.collegeconfidential.com/archive/index.php/t-197040.html&lt;/a&gt;&lt;/p>

5

<p>For 15</p>

<p>1) 2x - 5y = 8 throw 5y on the other side
2x = 8 + 5y divide by two
x = 4 + 2.5y</p>

<p>Plug into second equation</p>

<p>4x + ky = 17
4(4 +2.5y) + ky = 17
16 + 10y + ky = 17
16 + y(10 + k) = 17
y(10 + k) = 1
y = 1/(10 + k)</p>

<p>What value of k will make y undefined?... exactly, -10 will make y = 1/0 which is undefined, therefore, k = -10 would make a no solution</p>

6

<p>You don't need to list the combinations.</p>

<p>There's a 2/4 chance of placing one of them into an X building, because that's two buildings out of four. Then there's 1/3 chance of putting the remaining one into another X building, because there's only one more X building left out of the three unoccupied ones. So then you multiply 2/4 and 1/3 = 1/6.</p>

7

<p>Yeah I know, I was just doing it so he/she could get it better</p>