This question is incomplete- no bar chart was included.
The complete question was gotten from google.
A newspaper used a chart resembling the one to the right to illustrate the rising amounts that a video rental company spends to provide streaming content online. Is this the bar chart of a categorical variable, or is it a timeplot that uses bars to show the data?
Open the attachment below to see the bar chart'
Answer:
A newspaper used a chart resembling the one to the right to illustrate the rising amounts that a video rental company spends to provide streaming content online. Is this the bar chart of a categorical variable, or is it a time plot that uses bars to show the data? It is a time plot that uses bars to show the data; because it graphs a series of data recorded over time; presenting the values in sequential order.
Step-by-step explanation:
The chart used in the newspaper to illustrate the rising amounts that a video rental company spends to provide streaming content online is a time plot that uses bars to show the data; because it graphs a series of data recorded over time; presenting the values in sequential order.
Answer:
The area of a circle is equal to pi times the radius squared so that means the radius of this circle is the square root of. 25!that means the radius is 5
Answer:
Option B.
Step-by-step explanation:
It is given that a delivery truck driver charges a fixed base price of $6 for 2 miles.
After 2 miles, he charges an additional $2 for every mile and after 6 miles, he charges an additional $4 for every mile.
We need to describe the cost of the delivery truck between 1 mile and 2 miles.
In the given graph x-axis represents the distance in miles and y-axis represents the cost in dollars.
From the given graph it is clear that the value of function is constant between x=1 and x=2.
It means the cost of the delivery truck between 1 mile and 2 miles is constant.
Therefore, the correct option is B.
Adding 2 to each value of the random variable 
makes a new random variable 
. Its mean would be
![E[X+2]=E[X]+E[2]=E[X]+2](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3DE%5BX%5D%2BE%5B2%5D%3DE%5BX%5D%2B2)
since expectation is linear, and the expected value of a constant is that constant. ![E[X]](https://tex.z-dn.net/?f=E%5BX%5D)
is the mean of , so the new mean would be
![E[X+2]=10+2=12](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3D10%2B2%3D12)
The variance of a random variable is
![V[X]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
so the variance of would be
![V[X+2]=E[(X+2)^2]-E[X+2]^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5B%28X%2B2%29%5E2%5D-E%5BX%2B2%5D%5E2)
We already know ![E[X+2]=12](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3D12)
, so simplifying above, we get
![V[X+2]=E[X^2+4X+4]-12^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5BX%5E2%2B4X%2B4%5D-12%5E2)
![V[X+2]=E[X^2]+4E[X]+4-12^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5BX%5E2%5D%2B4E%5BX%5D%2B4-12%5E2)
![V[X+2]=(V[X]+E[X]^2)+4E[X]-140](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3D%28V%5BX%5D%2BE%5BX%5D%5E2%29%2B4E%5BX%5D-140)
Standard deviation is the square root of variance, so ![V[X]=3^2=9](https://tex.z-dn.net/?f=V%5BX%5D%3D3%5E2%3D9)
.
![\implies V[X+2]=(9+10^2)+4(10)-140=9](https://tex.z-dn.net/?f=%5Cimplies%20V%5BX%2B2%5D%3D%289%2B10%5E2%29%2B4%2810%29-140%3D9)
so the standard deviation remains unchanged at 3.
NB: More generally, the variance of 
for 
is
![V[aX+b]=a^2V[X]+b^2V[1]](https://tex.z-dn.net/?f=V%5BaX%2Bb%5D%3Da%5E2V%5BX%5D%2Bb%5E2V%5B1%5D)
but the variance of a constant is 0. In this case, 
, so we're left with ![V[X+2]=V[X]](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DV%5BX%5D)
, as expected.
(-4,-5)(-4,6)(3,8)
all y signs become opposite!:)
