HELPPPP! Math wizards over here :)

Test Preparation SAT Preparation
1

<p>Okay, I don't know how to get an image up to this site, so I am going to give you the book and whatnot.</p>

<p>It's in BB 2nd edition.
Test 4
Section 6
Question 17 & 18</p>

<p>Page 476 #17
(table needed)
A merchant sells 3 types of clocks that chime as indicated by the check marks in the table above. what is the total number of chimes of the inventory of clocks in the 90 minute period from 7:15 to 8:45</p>

<p>Page 476 #18
If 5 cards A B C D E are placed in a row so that C is never at either end, how many different arrangements are possible?</p>

2

<p>I don’t have the book right now, so I’ll elaborate on the third question only.</p>

<p>5 cards.
A B C D E.</p>

<p>You want card C to be in the 2nd, 3rd, and 4th, meaning that you don’t want it to be at either end.</p>

<p>For the first spot, how many different possibilities do you have? You can choose A, B, D, or E for the first spot. OK, that’s (4). Let’s assume you choose A.</p>

<p>For the second spot, you can choose B, D, C (You can’t choose A anymore, since you chose it for the first spot, and remember, now you can choose C because this isn’t either end. Also, you need to save one card for the end). That gives us another (3) choices. Let’s assume you choose B.</p>

<p>For the third spot, you can choose D, C, or E, following the same concept. That gives us another (3) choices. Let’s assume you chose C.</p>

<p>For the forth spot, you can choose D or E. That’s two options. (2)</p>

<p>And the last spot, you can only choose your last card, which is E, so that gives us 1.</p>

<p>You have a total of 72 possibilities for the card arrangements, according to the conditions listed.</p>

3

<p>i remember that 2nd question, it will be very difficult for me to explain it though :S :S try asking ur math teacher or any of ur frinds who got it rite</p>

4

<p>18) There are 5 positions to be filled: A B C D E
Let these dashes represent the positions: _ _ _ _ _
Fix A at one end and B at the other end. The remaining 3 cards can be arranged in 3!=6 ways (CDE, CED, DCE, DEC, ECD, EDC)</p>

<p>Number of ways in which A,B,D, or E taken 2 at a time can be arranged at the end= 4P2=4!/2!= 12.</p>

<p>In each of these 12 possibilities, the remaining cards can permute in 6 ways (see above)</p>

<p>So, total number of arrangements = 12 X 6 = 72</p>

<p>Hope its clear.</p>

5

<p>17) The first clock chimes n times every nth hour. Between 7:15 and 8:45, there is only 1 complete hour i.e., 8:00. The clock chimes 8 times at 8:00. It also chimes once every half hour. So, it chimes once at 7:30 and once at 8:30. Total no of chimes= 8+1+1=10.
There are 10 of these clocks. So total chimes by all 10 clocks= 10X10=100</p>

<p>Similarly, 1 clock B chimes 8 times and 5 clock B’s chime 8X5= 40 times.</p>

<p>Likewise, 3 clock C’s chime 9 times.</p>

<p>Total number of chimes=100+40+9=149.</p>

6

<p>thank you all (:</p>

7

<p>Here you could understand this easily first you have to see how many possible permutations for C</p>

<p>A-B-C-D-E</p>

<p>Here C can either go in 3 places so its 3!</p>

<p>–C–
-C—
—C-</p>

<p>Here you can see it visually by this:</p>

<p>ABCDE
ACBDE
ABDCE
ADCBE
ADBCE
ACDBE</p>

<p>Now you took 1 spot of possible permutations for C we have 4 spots left to occupy plus the permutations of C</p>

<p>then you have</p>

<p>4 * 3 * 2 * 1 * 2 = 72; </p>

<p>I recommend watching khanacademy.com on probability its awesome it will make you understand it perfectly.</p>

<p>4 * 3 * 2 * 1 =</p>