Final answer:
To calculate the number of bacteria present at 5:30 pm with an initial 15,000 at 3 pm and a doubling time of 1 1/2 hours, we find that there is one whole and one-third of a doubling period in the 2.5 hours from 3 pm to 5:30 pm. The population reaches approximately 47,622 bacteria by 5:30 pm.
Step-by-step explanation:
The subject of the question is Mathematics, specifically, it pertains to exponential growth which is a concept that is part of the high school curriculum. The question asks how many bacteria will be present at 5:30 pm if there are 15,000 bacteria at 3 pm and the colony has a doubling time of 1 1/2 or 1.5 hours.
Since the doubling time is 1.5 hours, from 3 pm to 5:30 pm, a total of 2.5 hours has passed. This time period includes one complete doubling time of 1.5 hours and an additional hour, resulting in one whole and one-third of a doubling time. To find the total number of bacteria at 5:30 pm, we perform the following calculation:
- After 1.5 hours (4:30 pm), the number of bacteria doubles from 15,000 to 30,000.
- After another 1 hour (5:30 pm), we have passed only 2/3 of the next doubling time. To calculate this partial doubling, we raise 2 to the power of 2/3 (since the bacteria double every 1.5 hours) which gives us approximately 1.5874. So, we multiply 30,000 bacteria by 1.5874.
The total number of bacteria present at 5:30 pm is approximately 47,622.