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ohaa [14]
3 years ago
12

lamont read 313.5 pages of his book in 5 hours 30 minutes. if he reads at a constant rate, hiw many pages of his book did lamont

read in 1 hour?
Mathematics
2 answers:
Rzqust [24]3 years ago
5 0
Lets find out how many pages he can read in a minute:

313.5 pages in (60*5)+30 = x minutes, x = 330 minutes

313.5/330 = .95 pages a minute

1 hour is 60 minutes

.95 * 60 = 57 pages in an hour.

Edit:
cross multiply to find how many pages read in an hour

313.5 pages                                   x pages
--------                                       =    -----------------
5 hours 30 mins (330 mins)            60 mins


(313.5*60)/330 = 57
S_A_V [24]3 years ago
3 0
57 pages per hour I think
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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)

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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}}

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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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