Question

Population is 10 minutes. at time t = 110 minutes, the bacterial the doubling period of a bacterial 50000. population was find the size of the baterial what was the initial population at time t = 0? please enter a whole number population after 3 hours. please enter a whole number

A) 800,000
B) 1,000,000
C) 1,250,000
D) 1,500,000

Answer 1

The initial population at t = 0 is
`\(50000 * 2^(-11)\)`, and after 3 hours, the bacterial population is 6,400,000.

C) 1,250,000.

Step-by-step explanation:

The initial population (
`\(P_0\)`) of the bacteria can be found using the formula
`\(P(t) = P_0 * 2^((t/d))\)`, where t is the time elapsed, d is the doubling period, and P(t) is the population at time t. Given that d = 10 minutes and
`\(P(110) = 50000 * 2^((110/10))\)`, we can solve for
`\(P_0\)`:

`\[ 50000 * 2^((110/10)) = P_0 * 2^((0/10)) \]`

Simplifying, we find
`\(P_0 = 50000 * 2^(-11)\)`. Now, to find the population after 3 hours (t = 180 minutes), we substitute t = 180 into the formula:

`\[ P(180) = P_0 * 2^((180/10)) \]`

Substituting the value of
`\(P_0\)` we found earlier, we calculate
`\(P(180) = 50000 * 2^(-11) * 2^(18)\)`.

`\(P(180) = 50000 * 2^7\)`

`\(50000 * 128 = 6,400,000\)`.

Therefore, the final population after 3 hours is 6,400,000.

In conclusion, the initial population
`(\(P_0\))` at t = 0 is
`\(50000 * 2^(-11)\)`, and the population after 3 hours is 6,400,000.

Answer 2

To determine the initial population and final population after 3 hours of bacterial growth, the exponential growth formula is used with a doubling period of 10 minutes. However, without the exact number of doubling periods for the provided options, the choice among A, B, C, and D cannot be confirmed.

Step-by-step explanation:

To solve for the initial population of bacteria at time t = 0 and the population size after 3 hours, we use the formula for exponential growth: final population = initial population × (2 ^(number of doubling periods)). Since the doubling period is every 10 minutes, we can find the number of doubling periods that occur in 110 minutes and in 3 hours (which is 180 minutes).

In the case given, the population at t = 110 minutes is 50,000 and we are given that this is after several doubling periods. To find the initial population, we divide 50,000 by 2 raised to the power of the number of doubling periods (which is 110 minutes / 10 minutes per period = 11 periods). So, the initial population would be 50,000 / 2^11.

For the size of the population after 3 hours, we need to calculate the number of doubling periods in 180 minutes (which is 180 / 10 = 18 doubling periods). Starting with the initial population we just calculated, we multiply it by 2^18 to get the population after 3 hours.

However, in this particular question, you've provided multiple-choice answers (A to D). The initial population and the population after 3 hours must be reassessed using the given options and counting the number of doubling periods accurately. Since the exact number of doubling periods for each figure is not provided here, we cannot confirm the correct choice.