Answer:
Ã=7/6
Step-by-step explanation:
<em>2/3(x -6) -y </em>
<h3><em>put the value of x= 9 and y=5/6 </em></h3>
=2/3(9-6)-5/6
=2/3×3-5/6
=2/1-5/6
=12-5/6
=7/6
<em>hey mate hope it's help you</em>......
144 inches. 12 feet in one hour
12 inches per feet
12feet x 12 inches per foot= 144 inches
Answer:
3, in both a), b)
Step-by-step explanation:
a) The slope of the line tangent to the curve that passes through the point (2,-10) is equal to the derivative of p at x=2.
Using differentiation rules (power rule and sum rule), the derivative of p(x) for any x is
. In particular, the value we are looking for is
.
If you would like to compute the equation of the tangent line, we can use the point-slope equation to get 
b) The instantaneus rate of change is also equal to the derivative of P at the point x=2, that is, P'(2). This is equal to
.
Answer:
80 degrees
Step-by-step explanatibecause its congruent to a suplementary angle
Answer:
![E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}](https://tex.z-dn.net/?f=E%28X%29%3D%20n%20%5Cint_%7B0%7D%5E1%20x%5En%20dx%20%3D%20n%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7D-%20%5Cfrac%7B0%7D%7Bn%2B1%7D%5D%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)
Step-by-step explanation:
A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".
We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).
If we select a value
we want this:

And we can express this like that:
for each possible i
We assume that the random variable
are independent and
from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

And then cumulative distribution would be expressed like this:



For each value
we can find the dendity function like this:

So then we have the pdf defined, and given by:
and 0 for other case
And now we can find the expected value for the random variable X like this:

![E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}](https://tex.z-dn.net/?f=E%28X%29%3D%20n%20%5Cint_%7B0%7D%5E1%20x%5En%20dx%20%3D%20n%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7D-%20%5Cfrac%7B0%7D%7Bn%2B1%7D%5D%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)