3(2c+d)-4(c-d)+d^2
c=1 , d=3
Use the substitution method
3(2*1+3)-4(1-3)+3^2
Multiply the brackets
3(2+3)-4+12+9
6+9-4+12+9
15-4+12+9
11+21
=32
Answer: 32
Answer:
15
Step-by-step explanation:
To solve this, divide 192 by 8:
192/9 = 24
Subtract 9 from 24:
24 - 9 = 15
Therefore, 8 times the sum of 9 and 15 is 192
Answer:
see explanation
Step-by-step explanation:
a
f(0) means find the value of y when x = 0
That is f(0) = 1 ← the point (0, 1) on the graph
b
When f(x) = - 3 means what are the values of x corresponding to y = - 3
From the graph when y = - 3 there are 2 corresponding values of x, that is
x = - 2 or x = + 2
The solution to f(x) = - 3 is x = ± 2
Answer:
35%
Step-by-step explanation:
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)
.
