Mathematics: Ascientist is testing a new antibiotic by applying the antibiotic to a colony of 10,000 bacteria. the number of bacteria decreases by 75% every two hours. how many hours will it take for the bacteria colony to decrease to 1000? round your answer to the nearest tenth of an hour.

7.5 because 6 + 9 = 15, and 15 /2 = 7.5...

Ascientist is testing a new antibiotic by applying

Ответ:

sarahholt517

10.03.2022

Mathematics

3.3 hours.

Step-by-step explanation:

We have been given that a scientist is testing a new antibiotic by applying the antibiotic to a colony of 10,000 bacteria. The number of bacteria decreases by 75% every two hours.

Since number of bacteria is decreasing exponentially, so we will use exponential decay function.

y=a*b^x , where,

a= Initial value,

b = For decay b is in form (1-r), where r is rate in decimal form.

Let us convert our given rate in decimal form.

$75\%=\frac{75}{100}=0.75$

As number of bacteria is decreasing every 2 hours, so number of bacteria decreased in 1 hour will be x/2.

Upon substituting our given values in above formula we will get,

$y=10,000(1-0.75)^{\frac{x}{2}}$

To find the number of hours it will take to for the bacteria colony to decrease to 1000, we will substitute y = 1,000 in our equation.

$1,000=10,000(0.25)^{\frac{x}{2}}$

Let us divide both sides of our equation by 10,000.

$\frac{1,000}{10,000}=\frac{10,000(0.25)^{\frac{x}{2}}}{10,000}$

$0.1=(0.25)^{\frac{x}{2}}$

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

$ln(0.1)=ln((0.25)^{\frac{x}{2}})$

$ln(0.1)=\frac{x}{2}*ln(0.25)$

$-2.302585=\frac{x}{2}*-1.386294$

$x=\frac{-2.302585}{-1.386294}*2$

x=1.660964*2

$x=3.3219\approx 3.3$

Therefore, it will take 3.3 hours for the bacteria colony to decrease to 1000.

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Ascientist is testing a new antibiotic by applying the antibiotic to a colony of 10,000 bacteria. the number of bacteria decreases by 75% every two hours. how many hours will it take for the bacteria colony to decrease to 1000? round your answer to the nearest tenth of an hour.

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