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3 years ago

11

An average person's heart beats about 103,680 times a day. Write and solve an equation to find about how many times the average

person's heart beats in one minute.

1 answer:

Damm [24]3 years ago

8 0

<span>It should be noted that in each day there are 24 hours which in turn is comprised of 60 minutes each. The way to answer this item to properly do the conversion and the dimensional analysis.
1. The equation that would be used in order to find out how many beats in one minute does an average human's heart do is,
x = (103,680 times/day)(1 day/24 hours)(1 hour/60 minutes) Simplifying,
x = 72 times / minute</span>

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Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

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Solving by separation of variables, then integrating both sides, we have that:

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$\int \frac{dP}{P} = \int r dt$

$\ln{P} = rt + K$

Applying the exponential to both sides:

$P(t) = Ke^{rt}$

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This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

So

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

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