Answer:
1. x=2 2.
y=1
Step-by-step explanation:
Im sorry I TRIED :(
Answer:
Step-by-step explanation:
Given the approximate demand function of night drink expressed as;
p^2+200q^2=177,
p is the price (in dollars) and;
q is the quantity demanded (in thousands).
Given
p = $7
q = 800
Required
dq/dp
Differentiating the function implicitly with respect to p shown;
2p + 400d dq/dp = 0
400q dq/dp = -2p
200qdq/dp = -p
dq/dp = -p/100q
substitute p and q into the resulting equation;
dq/dp = -7/100(800)
dq/dp = -7/80000
dq/dp = -0.0000875
This means that the rate of change of quantity demanded with respect to the price is -0.0000875
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)
.
Answer:
8
Step-by-step explanation:
160 divided by 20 hours = 8.
David has a 8 hourly wage.
