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In a lab experiment, a population of 400 bacteria is able to triple every hour. Which equation matches the number of bacteria in the population after 2 hours?

B=400(3)(3)

B=400(3)(3)(3)

B=3(1+400)^(2)

B=3(400)^(2)

In a lab experiment, a population of $400$ bacteria is able to triple every hour. Which equation matches the number of bacteria in the population after $2$ hours? $\newline$ $B=400(3)(3)$ $\newline$ $B=400(3)(3)(3)$ $\newline$ $B=3(1+400)^{2}$ $\newline$ $B=3(400)^{2}$

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Find derivatives of sine and cosine functions

Full solution

Q. In a lab experiment, a population of

400

bacteria is able to triple every hour. Which equation matches the number of bacteria in the population after

2

hours?

\newline

B=400(3)(3)

\newline

B=400(3)(3)(3)

\newline

B=3(1+400)^{2}

\newline

B=3(400)^{2}

Define Initial Population: Let's define the initial population of bacteria as $P_0$ . According to the problem, $P_0 = 400$ . The population triples every hour, so after one hour, the population is $3 \times P_0$ . After two hours, the population would be $3$ times the amount after one hour, which is $3 \times (3 \times P_0)$ .
Calculate Population Growth: Now, let's write the equation for the population after $2$ hours. We can denote the population after $2$ hours as $B$ . So, $B = 3 \times (3 \times P_0) = 3^2 \times P_0$ .
Write Population Equation: Substitute the initial population $P_0 = 400$ into the equation. So, $B = 3^2 \times 400 = 9 \times 400$ .
Substitute and Solve: Now, perform the multiplication to find the population after $2$ hours. $B = 9 \times 400 = 3600$ .

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