How do I solve?

Question

1. If the population of Paris is 8 million in 2000 and the decay rate is 6% annually, what is the expected population in 2007?

2. A colony of 5,000 bacteria is decreasing at 12% per hour. How many bacteria will there be in 24 hours?

Answer 1

1. It would be 8000000*(.94)^7=5187821--rounded to the nearest whole number.

2. 5000*(.88)^24=233--rounded to the nearest bacteria.

Answer 2

1. A=P(1- R/100)^n
=>A = 8*10^6 (1- 6/100)^7
=>A=5.18782075 × 10^12 = population of paris in 2007

2. similarly,
=>A=P(1- R/100)^n
=> A= 5000(1- 12/100)^24
=>A= 2.32570237 × 10^25 = amount of bacteria after 24 hours

Answer 3

"decay rate is 6% annually" means you lose 6% of the population each year, which leaves you with 94% of the previous population. So (this year's population) = (.94)*(last year's population). As the years go by, just keep multiplying by (.94) to get the new year's population. By 2007 you will have multiplied by (.94)^7 (to the 7th power).

Same principle with the bacteria - (1 - .12) = (.88), you have 88% this hour of what you had last hour. Multiply by (.88) every hour. After 24 hours, you've multipled 5000 by (.88)^24.

Answer 4

1. 464000000

2. -9400