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WINSTONCH [101]
3 years ago
6

A newborn baby's weight is recorded each month, from the day of birth until 12 months. One baby's weights, in pounds, are record

ed below:
7.20, 9.15, 11.01, 12.56, 14.04, 15.43, 16.52, 17.68, 18.77, 19.69, 20.22, 21.05, 21.90


What is the average amount of monthly weight gain, rounded to the nearest hundredth of a pound, each month?
A) 1.13
B) 15.79
C) 1.23
D) 16.52
Mathematics
2 answers:
Goryan [66]3 years ago
7 0
It can either be  A or C<span />
Nimfa-mama [501]3 years ago
3 0

Answer:

its C

Step-by-step explanation:

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5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l
sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

A(t)=100-80e^{-0.02t}

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})

Then we can rearrange and integrate

\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50}  \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}

Then we have the model of A(t) like

A(t)=100-80e^{-0.02t}

(b) The time at which the amount of salt reaches 50 lb is

A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

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