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scZoUnD [109]
3 years ago
9

A certain monthly magazine has both print and online subscribers. Print subscribers are people who pay to have the magazine phys

ically delivered to them each month. Online subscribers are people who pay to have access to the electronic version of the magazine. The editors of the magazine want to study how online subscribers feel about the design of the electronic version, and they will gather data from a sample. Which of the following is a sample of the population of interest?
a. 50 subscribers to the magazine
b. 50 print subscribers
c. 50 online subscribers
d. 50 people who buy a copy of the magazine at a newsstand
e. 50 people who find an article in the magazine while searching online
Mathematics
1 answer:
pentagon [3]3 years ago
4 0

<em>The sample population of interest is </em><em>50 online subscribers.</em>

<em>Sample population of interest</em> is said to be the <u>total number of people </u>one is considering for a particular study. In this question, online subscribers. This means the total people being considered for making the particular study.

to learn more, visit brainly.com/question/16720016

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-2m(m+n-4)+5(-2m+2n)+n(m+4n-5)
charle [14.2K]
Hello! And thank you for your question!

First we are going to expand the equation:

<span>−2<span>m^<span><span>​2</span><span>​​</span></span></span>−2mn+8m−10m+10n+nm+4<span>n<span><span>​^2</span><span>​​</span></span></span>−5n

Then we are going to combine like terms:

</span><span><span>−2<span>m<span><span>​^2</span><span>​​</span></span></span>+(−2mn+mn)+(8m−10m)+(10n−5n)+4<span>n<span><span>​^2</span><span>​​</span></span></span></span>
</span>
Then finally, simplify:

−2<span>m<span><span>​^2</span><span>​​</span></span></span>−mn−2m+5n+4<span>n<span>​^<span>2

Final Answer:
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8 0
4 years ago
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Find a power series representation for the function and determine the interval of convergence ln(5 - x
Andrew [12]
To find the power series representation of In(5 - x), we recall that the power series representation of a function of the form:
\frac{a}{1-r} =\sum\limits^\infty_{n=0}ar^n
provided |r| < 1
Also recall that
\int{\frac{a}{1-r}\,dx} =\int{\sum\limits^\infty_{n=0}ar^n\,dx}
Notice that
\ln{(5-x)}=-\int {\frac{1}{5-x}\,dx}
To get the power series of
\frac{1}{5-x}= (\frac{1}{5})  \frac{1}{1- \frac{x}{5} }  \\ =\sum\limits^\infty_{n=0}( \frac{1}{5}) (\frac{x}{5})^n=(\frac{1}{5}) (\frac{x}{5})+(\frac{1}{5}) (\frac{x^2}{25})+(\frac{1}{5}) (\frac{x^3}{125})+ . . . \\ =\frac{x}{25}+\frac{x^2}{125}+ \frac{x^3}{625}+ . . .
Therefore, the power series representation of
\ln{(5-x)}=-\int {\frac{1}{5-x}\,dx} \\ =-\int{(\frac{x}{25}+\frac{x^2}{125}+ \frac{x^3}{625}+ . . . ),dx} \\ =C-\frac{x^2}{50}-\frac{x^3}{375}- \frac{x^4}{2500}- . . .
When x = 0: C = ln 5
Therefore,
\ln{(5-x)}=\ln{5}-\frac{x^2}{50}-\frac{x^3}{375}- \frac{x^4}{2500}- . . .

The radius of convergence is given by |r| < 1.
Here,
r= \frac{x}{5}
Therefore, radius of convergence is
| \frac{x}{5}| \ \textless \ 1 \\ |x|\ \textless \ 5
4 0
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