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Lena [83]
3 years ago
9

PLZZZZZZ HELPPPPPP QUESTION 2 ATTACHMENT BELOW

Mathematics
1 answer:
Harrizon [31]3 years ago
7 0
Hello! So, you have a ½ chance of landing on black. To find the possibility that it will land on black 3 times, we multiply fractions, because when it comes to getting a combination, chances lower. ½ * ½ * ½ is 1/8. There is 1/8 of a chance that the spinner will land on black 3 times. The answer is D: 1/8.
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It is d y-1x2-Bc+29 I say that bc it says that in the answer up above
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A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t
sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

a)

\frac{dy}{dt}\ \alpha\  y(1-y)

\frac{dy}{dt}=ky(1-y)

where k is the constant of proportionality, dy/dt =  rate at which the rumor spreads

b)

\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}}  =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\

y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days

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3 years ago
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(f-g)x= -x²+13x+13

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Hope that helps!
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