Answer: She can make two 3/4 yards long of oak molding. She will have half of the second one left over.
Step-by-step explanation:
Can you please explain? Or maybe continue your question please?
The expected length of code for one encoded symbol is

where
is the probability of picking the letter
, and
is the length of code needed to encode
.
is given to us, and we have

so that we expect a contribution of

bits to the code per encoded letter. For a string of length
, we would then expect
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have

so that the variance for the length such a string is

"squared" bits per encoded letter. For a string of length
, we would get
.
The first three terms of sequence are 9 , 6 , 3
<em><u>Solution:</u></em>
Given the recursive function f(n) = f(n - 1) - 3
Where f(1) = 9
To find: First three terms of sequence
Substitute n = 2 , n = 3 and n = 4 in given recursive function
When n = 2
f(n) = f(n - 1) - 3
f(2) = f(2 - 1) - 3
f(2) = f(1) - 3
f(2) = 9 - 3 = 6
f(2) = 6
Thus second term is 6
When n = 3
f(3) = f( 3 - 1) - 3
f(3) = f(2) - 3
f(3) = 6 - 3 = 3
f(3) = 3
Thus the third term is 3
When n = 4
f(4) = f( 4 - 1) - 3
f(4) = f(3) - 3
f(4) = 3 - 3
f(4) = 0
Thus the fourth term is 0
Thus first three terms of sequence are 9 , 6 , 3
(- 1 + √3)/2
rationalise the denominator by multiplying the numerator/denominator by the conjugate of the denominator.
the conjugate of 1 + √3 is 1 - √3
(1 - √3)/(1 + √3)(1 - √3) = (1 - √3)/( 1² - (√3)²) = (1 - √3)/(1 - 3) = (1 - √3)/(- 2)
multiply numerator/denominator by (- 1)
=(-1 +√3)/2