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MrRa [10]
3 years ago
13

A bacteria population experiences continuous growth. If there were 2,000 bacteria present at the start of an experiment, how lon

g did it take for there to be 6,000 bacteria if the growth rate was 7% per hour
about 0.06 hours

about 3.76 hours

about 6.82 hours

about 15.69 hours
Mathematics
1 answer:
allochka39001 [22]3 years ago
3 0

Answer:

D. About 15.69 hours.

Step-by-step explanation:

Let x be the number of hours.

We have been given that there  were 2,000 bacteria present at the start of an experiment and the growth rate was 7% per hour.

Since number of bacteria is growing exponentially, so we will use exponential growth function to solve our given problem.

The continuous exponential growth formula is in form: y=e^{kx}, where,

e= mathematical constant,

k = Growth rate in decimal form.

Let us convert our given rate in decimal form.

7\%=\frac{7}{100}=0.07

Upon substituting our given values we will get exponential function for bacteria growth as: y=2,000*e^{0.07x}, where, y represents number of bacteria after x hours.

Since we need to figure out the number of hours it will take for there to be 6,000 bacteria, so we will substitute y= 6,000 in our function.

6,000=2,000*e^{0.07x}

Let us divide both sides of our equation by 2,000.

\frac{6,000}{2,000}=\frac{2,000*e^{0.07x}}{2,000}

3=e^{0.07x}

Let us take natural log of both sides of our equation.

\text{ln}(3)=ln(e^{0.07x})

\text{ln}(3)=0.07x

1.0986122=0.07x  

x=\frac{1.0986122}{0.07}  

x=15.69446\approx 15.69  

Therefore, it will take about 15.69 hours to be 6,000 bacteria and option D is the correct choice.




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