Friday

April 18, 2014

April 18, 2014

Posted by **Jon** on Friday, February 22, 2008 at 2:03pm.

A)5/3

B)5/2

C)3/5

D)does not exist

I chose B

(3/5)/1 = .6

S = a1/1-r

1/1-.6

2.5

2)Use the Binomial Theorem to find the sixth term in the expansion of (m+2p)^7.

A)21m^2p^5

B)672m^2p^5

C)32m^2p^5

D)448mp^6

7/(7-k)!k!^m7-kpk

7/(7-5)!5!^m7-5p5

7*6*5*4*3/5*4*3*2*1^m2p5

2520/120

21m^2p^5

3)How many four-digit numerical codes can be created if no digit may be repeated?

A)10,000

B)24

C)3024

D)5040

I chose A

10*10*10*10

10,000

- Algebra -
**Reiny**, Friday, February 22, 2008 at 3:19pm#2 You forgot that 2 is also raised to the fifth.

C(7,5)(m^2)(2p(^5

= 21(2^5)m^2p^5

=672m^2p^5

#3. Digits may NOT be repeated, and probably cannot start with a zero.

so 9x9x8x7 = 4536, which is none of their choices.

if zeros would be allowed in the first position, then the answer would be

1-x9x8x7 = 5040 which is choice D

your choice A would include zeros, and repeating digits, even 0000,and 6666.

I would guess that D is the choice they are after.

#1 is ok

- Algebra -
**Reiny**, Friday, February 22, 2008 at 3:20pmtypo in "1-x9x8x7 = 5040 which is choice D "

should say:

10x9x8x7 = 5040 which is choice D

- Algebra -

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