Answer:
x = 11, -1
Step-by-step explanation:
First, let's identify what the quadratic formula is:
x = [-b ± √(b² - 4(a)(c))] / 2
Our equation is written in standard form:
ax² + bx + c = 0
x² - 10x - 11 = 0
Let's plug in what we know.
x = [-(-10) ± √((-10)² - 4(1)(-11))] / 2
Evaluate the exponent.
x = [-(-10) ± √(100 - 4(1)(-11))] / 2
Simplify the negatives.
x = [10 ± √(100 - 4(1)(-11))] / 2
Multiply.
x = [10 ± √(100 + 44)] / 2
Simplify the parentheses.
x = [10 ± √(144)] / 2
Simplify the radical (√)
x = [10 ± 12] / 2
Evaluate the ±.
x = [10 + 12] / 2
x = [22] / 2
x = 11
or
x = [10 - 12] / 2
x = [-2] / 2
x = -1
Your answers are x = 11, -1
Hope this helps!
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 ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n)
.
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 ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n)
.
S=Years Susanna has played the piano
p=Years Patrick has played the piano
The expression for this question would be s(2)+4=p
Hope this helps! :)
Answer:
-15
Step-by-step explanation:
the new temperature since it went down
-12-3= -15
