How many bacteria would be present 15 hours after the experiment began if a set of bacteria begins with 20 and doubles every 2 hours?

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How many bacteria would be present 15 hours after the experiment began if a set of bacteria begins with 20 and doubles every 2 hours?

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Answer 1 · 2014-12-07T22:03:55

The answer is $3811$ $3811$ bacteria.

Using a function to describe exponential growth, we can say that

$A = A_{\circ} \cdot e^{k \cdot t}$ $A = A_@ * e^(k*t)$ , where

$A$ $A$ - is the amount we need to find out (in this case, the number of bacteria after 15 hours of growth);
$A_{\circ}$ $A_@$ - the initial number of bacteria;
$k$ $k$ - growth rate;
$t$ $t$ - time;

We were given $t$ $t$ = $15$ $15$ hours and $A_{\circ}$ $A_@$ = $20$ $20$ bacteria; however, both $k$ $k$ and $A$ $A$ need to be determined.

We will determine $k$ $k$ by using the fact that the number of bacteria doubles every two hours - this means that after the first 2 hours, we will have 40 bacteria. So,

$40 = 20 \cdot e^{k \cdot 2}$ $40 = 20 * e^(k*2)$ , which gives us a $k$ $k$ = $0.35$ $0.35$ .

Therefore, $A = 20 \cdot e^{0.35 \cdot 15} = 3811$ $A = 20 * e^(0.35 * 15) = 3811$ bacteria.

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How many bacteria would be present 15 hours after the experiment began if a set of bacteria begins with 20 and doubles every 2 hours?

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