The question is defective, or at least is trying to lead you down the primrose path.
The function is linear, so the rate of change is the same no matter what interval (section) of it you're looking at.
The "rate of change" is just the slope of the function in the section. That's
(change in f(x) ) / (change in 'x') between the ends of the section.
In Section A:Length of the section = (1 - 0) = 1f(1) = 5f(0) = 0change in the value of the function = (5 - 0) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
In Section B:Length of the section = (3 - 2) = 1 f(3) = 15f(2) = 10change in the value of the function = (15 - 10) = 5Rate of change = (change in the value of the function) / (size of the section) = 5/1 = 5
Part A:The average rate of change of each section is 5.
Part B:The average rate of change of Section B is equal to the average rate of change of Section A.
Explanation:The average rates of change in every section are equalbecause the function is linear, its graph is a straight line,and the rate of change is just the slope of the graph.
You would start by figuring out the distance the turtle swims every second by dividing 24 by 16.
24➗16=1.5
From this we can take the numbers from either T: 1, ___, 24, 45 or D: ___, 10, 16, ___ to figure out the blank spaces for the other using multiplication or division. Divide T by 1.5, or multiply D by 1.5 to find the other
T: 1, 15, 24, 45
D: 1.5, 10, 16, 30
A. so the 3 before the 3(r-5) represents the initial price, in this circumstance being 3
B. the R represents the rentals, for every R another video game has been rented
C. The 5 represents the money subtracted for every video game rented.
Answer:
It is a growth.
Step-by-step explanation:
If the number in parenthesis (exponential base) is greater than 1, then it is a growth. If it is between 1 and 0, then it is a decay.
I'm going to assume the joint density function is
a. In order for 
to be a proper probability density function, the integral over its support must be 1.
b. You get the marginal density 
by integrating the joint density over all possible values of 
:
c. We have
d. We have
and by definition of conditional probability,
e. We can find the expectation of 
using the marginal distribution found earlier.
![E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_0%5E1xf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac67%5Cint_0%5E1%282x%5E2%2Bx%29%5C%2C%5Cmathrm%20dx%3D%5Cboxed%7B%5Cfrac57%7D)
f. This part is cut off, but if you're supposed to find the expectation of , there are several ways to do so.
- Compute the marginal density of , then directly compute the expected value.
![\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac87)
- Compute the conditional density of given
, then use the law of total expectation.
The law of total expectation says
![E[Y]=E[E[Y\mid X]]](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5BE%5BY%5Cmid%20X%5D%5D)
We have
![E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}](https://tex.z-dn.net/?f=E%5BY%5Cmid%20X%3Dx%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_%7BY%5Cmid%20X%7D%28y%5Cmid%20x%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac%7B6x%2B4%7D%7B6x%2B3%7D%3D1%2B%5Cfrac1%7B6x%2B3%7D)
![\implies E[Y\mid X]=1+\dfrac1{6X+3}](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5Cmid%20X%5D%3D1%2B%5Cdfrac1%7B6X%2B3%7D)
This random variable is undefined only when 
which is outside the support of , so we have
![E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5Cleft%5B1%2B%5Cdfrac1%7B6X%2B3%7D%5Cright%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cleft%281%2B%5Cfrac1%7B6x%2B3%7D%5Cright%29f_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac87)
