A rumor spreads in a small town with population 20,000 people. The secret spreads throughout the town according to the function, where t represents time, in days: N(t)=20000/10+2490^.97t A.) How many people were in the group that started the rumor? B.) How many people have heard the rumor after 1 day? One week? C.) At this rate, how long will it take for 1950 people to hear the rumor?
\[N(t)=\frac{ 20000 }{ 10+2490e^{-0.97t} }\]
i should make a template for this one have seen this exact question over a dozen times in the past two years
A.) How many people were in the group that started the rumor? put \(t=0\) what do you get?
hint: \(e^0=1\)
It'll just be the exact same problem won't it? 10+2490?
that is in the denomnator
\[\frac{20,000}{10+2490}\] compute that one
8 people ?
20000/3000 which gives me a long decimal
\[N(t)=\frac{ 20000 }{ 10+2490e^{-0.97t} }\\ N(0)=\frac{ 20000 }{ 10+2490e^{0} }=\frac{20,000}{10+2490}\]
oh no dear, what is \(2490+10\) ?
Oh... Woops... Haha 20000/2500 8 c:
whew
B.) How many people have heard the rumor after 1 day? \[N(t)=\frac{ 20000 }{ 10+2490e^{-0.97t} }\\ N(1)=\frac{ 20000 }{ 10+2490e^{-.097} }=\text{ use a calculator}\]
B.) How many people have heard the rumor after 1 week put \(t=7\) compute \[N(7)=\frac{ 20000 }{ 10+2490e^{-.097\times 7} }=\text{ use a calculator}\]
oops i typed it in wrong in to wolf, let me redo the previous one
The second one is almost 16... 15.714
same answer though, not quite 9 in any case you can use the wolfram link i sent as a template if you do not want to plug this in to a calculator this would be for 7 days http://www.wolframalpha.com/input/?i=20000%2F%2810%2B2490e^%28-0.097*7%29%29
yeah that is what i got too
And that's B?
yes
C.) At this rate, how long will it take for 1950 people to hear the rumor? this is the only one that requires any real work any ideas?
would I do the same thing as the last one, but plug in 1950 for t?
oh no
this one says the ANSWER is 1950, what \(t\) did you plug in to get it?
\[N(t)=\frac{ 20000 }{ 10+2490e^{-0.97t} }=1950\] solve for \(t\)
Can yhu tell me what my first step would be?
multiply by the denominator to clear the fraction \[20,000=1950(10+e^{-.97t})\]
damn typo \[20,000=1950(10+2490e^{-97t})\]
then distribute teh \(1950\) on the right
or you can divide both sides by 1950 first either way
\[20,000= 19,500+4,855,500e^{-.97t}\]
;-;
i believe you
subtract 19500 from both sides
even i can do that without a calculator \[500=4,855,500e^{-.97t}\]
\[500=4,855,500e^{-.97}\] c:
divide by \(4,855,500\)
4 million divided by 500? Or 500 divided by the 4 million?
second way it is a tiny fraction \[\frac{1}{9771}\]
... 9771? I got 9711... ;c
now comes the only non algebra step \[\frac{1}{9771}=e^{-.97t}\]
that's cause you did it upside down
\[a=bx\\ \frac{a}{b}=x\]
ok now do you know what to do here?\[\frac{1}{9771}=e^{-.97t}\]
No ;-; And I keep getting 9711. ;c I flipped it around and got 1.long decimal ;C
"no" is a fine answer forget the decimal
you got \(9771\) because you divided 500 by 4855500
ack i mean that is what you did not do you divided backwards
it should be \[500\div 4855500\] and you should get the reciprocal of \(9771\) which is \(\frac{1}{9771}\) a very tiny decimal, lets leave it as a fraction
;c
...........................
Wait the decimal is right?
oops typo again!
\[\frac{1}{9711}=e^{-.97t}\\ \ln(\frac{1}{9711})=-.97t\]
and finally divide by \(-.97\) to find \(t\) i used this http://www.wolframalpha.com/input/?i=ln%281%2F9711%29%2F-.97
Aha! Ok... I'm caught back up. :D So yhu replaced the e with a natural log... And I would divide 1/9711 by -.97?
take the log of both sides i.e. write in logarithmic form one you have \[\frac{1}{9711}=e^{-.97t}\] you have to get \(t\) out of the exponent you do that via \[\ln(\frac{1}{9711})=-.97t\]
then simple algebra to get \(t\) namely divide both sides by \(-.97\)
i sent you the wolfram link with what i am pretty sure is the answer, typo free
gotta run, good luck
Ok n.n I kind of understand how this was worked out. c: Thank yhu Satellite ^.^
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