y=19x+9 would be the rule
Domain: the set of possible values of the independent variable or variables of a function.
Range of graph: The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values.
Answer:
<u>A. The probability that a Millennial is married is 0.089 or 8.9%.</u>
<u>B. The probability that a Baby Boomer is single, never married is 0.03 or 3%.</u>
<u>C. The probability that one person selected randomly (female or male) is married is 0.513 or 51.3% </u>
<u>D. The probability that someone who is living with a partner, but not married is a Generation X is 0.025 or 2.5%.</u>
Step-by-step explanation:
According to the information provided on the analysis table, we can answer the questions:
A. The probability that a Millennial is married is 0.089 or 8.9%.
B. The probability that a Baby Boomer is single, never married is 0.03 or 3%.
C. The probability that one person selected randomly (female or male) is married is 0.513 or 51.3% (Millennial 0.089 + Generation X 0.223 + Baby boomer 0.201)
D. The probability that someone who is living with a partner, but not married is a Generation X is 0.025 or 2.5%.
I assume
has counterclockwise orientation when viewed from above.
By Stokes' theorem,
![\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r%3D%5Ciint_S%28%5Cnabla%5Ctimes%5Cvec%20F%29%5Ccdot%5Cmathrm%20d%5Cvec%20S)
so we first compute the curl:
![\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20F%28x%2Cy%2Cz%29%3Dxy%5C%2C%5Cvec%5Cimath%2Byz%5C%2C%5Cvec%5Cjmath%2Bxz%5C%2C%5Cvec%20k)
![\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k](https://tex.z-dn.net/?f=%5Cimplies%5Cnabla%5Ctimes%5Cvec%20F%28x%2Cy%2Cz%29%3D-y%5C%2C%5Cvec%5Cimath-z%5C%2C%5Cvec%5Cjmath-x%5C%2C%5Cvec%20k)
Then parameterize
by
![\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20r%28u%2Cv%29%3D%5Ccos%20u%5Csin%20v%5C%2C%5Cvec%5Cimath%2B%5Csin%20u%5Csin%20v%5C%2C%5Cvec%5Cjmath%2B%5Ccos%5E2v%5C%2C%5Cvec%20k)
where the
-component is obtained from
![1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v](https://tex.z-dn.net/?f=1-%28%5Ccos%20u%5Csin%20v%29%5E2-%28%5Csin%20u%5Csin%20v%29%5E2%3D1-%5Csin%5E2v%3D%5Ccos%5E2v)
with
and
.
Take the normal vector to
to be
![\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20r_v%5Ctimes%5Cvec%20r_u%3D2%5Ccos%20u%5Ccos%20v%5Csin%5E2v%5C%2C%5Cvec%5Cimath%2B%5Csin%20u%5Csin%20v%5Csin%282v%29%5C%2C%5Cvec%5Cjmath%2B%5Ccos%20v%5Csin%20v%5C%2C%5Cvec%20k)
Then the line integral is equal in value to the surface integral,
![\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%28%5Cnabla%5Ctimes%5Cvec%20F%29%5Ccdot%5Cmathrm%20d%5Cvec%20S)
![=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_0%5E%7B%5Cpi%2F2%7D%28-%5Csin%20u%5Csin%20v%5C%2C%5Cvec%5Cimath-%5Ccos%5E2v%5C%2C%5Cvec%5Cjmath-%5Ccos%20u%5Csin%20v%5C%2C%5Cvec%20k%29%5Ccdot%28%5Cvec%20r_v%5Ctimes%5Cvec%20r_u%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle-%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Ccos%20v%5Csin%5E2v%28%5Ccos%20u%2B2%5Ccos%5E2v%5Csin%20u%2B%5Csin%282u%29%5Csin%20v%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Cboxed%7B-%5Cfrac%7B17%7D%7B20%7D%7D)
Answer:
The given statement is false.
Step-by-step explanation:
Circle graphs, or pie graphs, show information over a period of time.
This statement is false.
A circle graph, or a pie chart, helps in visualizing information and data. These graphs are used to show the results in a proportional manner.
The circle graph is divided in proportions that show the percentage of a certain answer.