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artcher [175]
3 years ago
10

A culture of bacteria obeys the law of uninhibited growth. if 500 bacteria are present initially, and there are 800 after 1 hour

, how many will be present after 5 hours? round to the nearest whole number.
Mathematics
2 answers:
lesantik [10]3 years ago
8 0
Since this culture of bacteria obeys the law of uninhibited growth. Given that 500 bacteria are present initially, and there are 800 after 1 hour, a growth rate of 0.470003629246 (k value) has been calculated. <span>5243 </span>will be present after 5 hours.
Alla [95]3 years ago
5 0
The law of uninhibited growth follows below formula:

A = Pe^(rt)

For your problem, we need to first look for the rate, given the initial and final number of bacteria..

A = Pe^(rt)
800 = (500)(e)^r(1)

(800/500) = (e)^r(1)
ln (8/5) = ln (e) ^ r
r = 0.47

substitute the value of r to get the number of bacteria after 5 hours:

A = Pe^(rt)
A = (500)(e)^(0.47)(5)

A = 5,242.78 or 5,243
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<h2>Hope it helps </h2>
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4 years ago
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vitfil [10]

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1

.

1

+

1

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(

x

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⋅

sin

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x

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7 0
3 years ago
I need help I don’t understand
aleksklad [387]
<h3><em>1) 0.7</em></h3><h3><em>2) 0.42</em></h3><h3><em>3) 0.25</em></h3><h3><em>4) 0.375</em></h3><h3><em>5) 0.33</em></h3><h3><em>6) 0.071</em></h3><h3><em>7) 0.775</em></h3><h3><em>8) 0.775</em></h3><h3><em>9) 0.682</em></h3><h3><em>10) 0.791</em></h3><h3><em>HOPE IT HELPS !!!</em></h3>

7 0
3 years ago
$10,000 at an annual rate of 7%, compounded semi-annually, for 2 years
forsale [732]

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8 0
3 years ago
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