Question

HELP PLEASE I WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!

A bacteria colony doubles every 40 minutes. If the colony contains 10 bacteria at noon, how many bacteria are present at 6 pm?

A.180

B.512

C.810

D.900

E.5120

Answer 1

Answer: E

noon to 6 pm is 6 hours which is 360 minutes

divide 360 by 40 to get the number of times it doubles which is 9 times so it’ll be 2 to the ninth power (2 is there since your doubling the number of bacteria)

multiply that by 10 since that was how many bacteria there was at the beginning

10 x 2^9 = 5120

Answer 2

5120 bacteria will be there at 6 PM.

This question is an example of geometric progression, in which a bacteria colony has growing population that is multiplying itself at constant rate. The geometric progression is defined by this expression:

$n (t) = n_{o} \cdot r^{\frac{t}{\tau} }$ (1)

Where:

• $n_{o}$ - Initial population, no unit.
• $r$ - Growth rate, no unit.
• $t$ - Time, in minutes.
• $\tau$ - Time factor, in minutes.

If we know that $n_{o} = 10$, $r = 2$, $\tau = 40\,min$ and $t = 360\,min$, then the population of bacteria at 6 PM is:

$n(360) = 10\cdot 2^{\frac{360}{40} }$

$n (360) = 5120$

5120 bacteria will be there at 6 PM.

We kindly invite you to check this question related to geometric progression: https://brainly.com/question/24540317