Answer:
The correct option is;
Use a scale factor of 2
Step-by-step explanation:
The parameters given are;
A = (1, -6)
B = (5, -6)
C = (6, -2)
D = (0, -2)
A'' = (1.5, 4)
B'' = (3.5, 4)
C'' = (4, 2)
D'' = ( 1, 2)
We note that the length of side AB in polygon ABCD = √((5 -1)² + (-6 - (-6))²) = 4
The length of side A''B'' in polygon A''B''C''D'' = √((3.5 -1.5)² + (4 - 4)²) = 2
Which gives;
AB/A''B'' = 4/2 = 2
Similarly;
The length of side BC in polygon ABCD = √((6 -5)² + (-2 - (-6))²) = √17
The length of side B''C'' in polygon A''B''C''D'' = √((4 -3.5)² + (2 - 4)²) = (√17)/2
Also we have;
The length of side CD in polygon ABCD = √((6 -0)² + (-2 - (-2))²) = 6
The length of side C''D'' in polygon A''B''C''D'' = √((4 -1)² + (2 - 2)²) = 3
For the side DA and D''A'', we have;
The length of side DA in polygon ABCD = √((1 -0)² + (-6 - (-2))²) = √17
The length of side D''A'' in polygon A''B''C''D'' = √((1.5 -1)² + (4 - 2)²) = (√17)/2
Therefore the Polygon A B C D can be obtained from polygon A''B''C''D'' by multiplying each side of polygon A''B''C''D'' by 2
The correct option is therefore;
Use a scale factor of 2.
Think of it this way: Lets add numbers in pairs, starting at the very outer 2 numbers (19 and 77) then go in by one and add the second number and the second to last (20 and 76), then (21 and 75) and so on. The sum of all of these pairs are all the same: 96. How many 96s will we have? Well since we're coming from each end toward the middle adding pairs we will have half the distance between 19 and 77, that is (77-19)/2 = 29. So we can actually just take 96*29 = 2784. This is the sum of all numbers between 19 and 77
Answer:
Yes
Step-by-step explanation:
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)
Answer:
Step-by-step explanation:
(C). A( - 4, - 1 ), B( 3, 0 ), C( - 7, 2 )