She would put 0.35 lbs of rice in each bag. Hope it helps!
The relation represented by the arrow diagram is {(-3, 4), (-1, 5), (0, 7), (2, 2), (5, 7)}.
Option: C.
<u>Step-by-step explanation:</u>
A function is a relation in which each input value(domain) results in one output value(range). It is represented diagrammatically using the mapping method.
It shows how each element of domain and range are paired. That is like a flowchart it shows the input values marking its corresponding output value.
In the given diagram,
The values given in the left are domain and values given in the right are range.
Thus, -3 marks to 4, then can be written as (-3,4).
Similarly,
-1 marks 5 = (-1,5).
0 marks 7= (0,7).
2 marks 2= (2,2).
5 marks 7 =(5,7).
⇒ The complete points sequence is {(-3, 4), (-1, 5), (0, 7), (2, 2), (5, 7)}.
Answer: T = 4
Step-by-step explanation:
1. Write all the variables down
P = 8 V = 2X N = 2 R = 2X X = 3
2. Since you know that X = 3 substitute it in to find V and R
V = 2X = 2(3) = 6
R = 2X = 2(3) = 6
3. Find PV
PV = P x V
= 8 x 6
= 48
4. Find NRT
NRT = N x R x T
= 2 x 6 x T
= 12 x T
= 12T
5. Find T
PV = NRT
48 = 12T
12T = 48
divide both sides by 12
T = 48 ÷ 12
T = 4
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)
.
