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Re: The value of (2^-14)+(2^-15)+(2^-16) + (2^-17) is how ... [#permalink]
14 Apr 2012, 13:50

3

This post received KUDOS

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andih wrote:

The value of (2^-14)+(2^-15)+(2^-16) + (2^-17) is how times the value of 2^-17?

A. 3/2

B. 5/2

C. 3

D. 4

E. 5

Original question reads: The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is how many times the value of 2^(-17)?

We need to find the value of: \(\frac{\frac{1}{5}*(2^{-14}+2^{-15}+2^{-16}+2^{-17})}{ 2^{-17}}=\frac{\frac{1}{5}*(\frac{1}{2^{14}}+\frac{1}{2^{15}}+\frac{1}{2^{16}}+\frac{1}{2^{17}})}{\frac{1}{2^{17}}}\).

Re: The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is [#permalink]
15 Apr 2012, 06:47

3

This post received KUDOS

1

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if you take 2^(-17) common in the numerator, you will have 2^(-17) { 8 + 4 + 2 + 1} which equals 15. This 15 cancels with 5 in the denominator and leaves {3} 2^-(17). Slightly quicker this way i feel.

Re: The value of (2^-14)+(2^-15)+(2^-16) + (2^-17) is how ... [#permalink]
09 Oct 2013, 19:41

Bunuel wrote:

andih wrote:

The value of (2^-14)+(2^-15)+(2^-16) + (2^-17) is how times the value of 2^-17?

A. 3/2

B. 5/2

C. 3

D. 4

E. 5

Original question reads: The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is how many times the value of 2^(-17)?

We need to find the value of: \(\frac{\frac{1}{5}*(2^{-14}+2^{-15}+2^{-16}+2^{-17})}{ 2^{-17}}=\frac{\frac{1}{5}*(\frac{1}{2^{14}}+\frac{1}{2^{15}}+\frac{1}{2^{16}}+\frac{1}{2^{17}})}{\frac{1}{2^{17}}}\).

Re: The value of (2^-14)+(2^-15)+(2^-16) + (2^-17) is how ... [#permalink]
09 Oct 2013, 20:15

runningguy wrote:

Bunuel wrote:

andih wrote:

The value of (2^-14)+(2^-15)+(2^-16) + (2^-17) is how times the value of 2^-17?

A. 3/2

B. 5/2

C. 3

D. 4

E. 5

Original question reads: The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is how many times the value of 2^(-17)?

We need to find the value of: \(\frac{\frac{1}{5}*(2^{-14}+2^{-15}+2^{-16}+2^{-17})}{ 2^{-17}}=\frac{\frac{1}{5}*(\frac{1}{2^{14}}+\frac{1}{2^{15}}+\frac{1}{2^{16}}+\frac{1}{2^{17}})}{\frac{1}{2^{17}}}\).

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