1. you deposit $1500 in an account that pays 5% interest yearly. how much money do you have after 6 years?
2. if i have $500 in my account after 4 years of investing at 2.5% per year, how much money did i start with?
3. a hypothetical strain of bacteria doubles every 5 minutes (exponential growth). one single bacterium was put in a sealed bottle at 9 a.m. and the bottle was filled at exactly 10 a.m. at what time was the bottle one-half full? think in terms of doubling time
4. the population of a city grows at a rate of 5% per year. the population in 1990 was 400,000. what would be the predicted current population? in what year would we predict the population to reach 1,000,000?
2. if i have $500 in my account after 4 years of investing at 2.5% per year, how much money did i start with?
3. a hypothetical strain of bacteria doubles every 5 minutes (exponential growth). one single bacterium was put in a sealed bottle at 9 a.m. and the bottle was filled at exactly 10 a.m. at what time was the bottle one-half full? think in terms of doubling time
4. the population of a city grows at a rate of 5% per year. the population in 1990 was 400,000. what would be the predicted current population? in what year would we predict the population to reach 1,000,000?
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1. you deposit $1500 in an account that pays 5% interest yearly. how much money do you have after 6 years?
After one year, the amount of money in the bank is:
1500*1.05.
After two years:
1500*1.05*1.05 = 1500*1.05^2.
Generally:
1500*10.5^t, where t is the time that has passed in years.
2. if i have $500 in my account after 4 years of investing at 2.5% per year, how much money did i start with?
using the same form as above:
a*1.025^t = 500.
a = 500/(1.025^t), where t=6 and a is the starting amount.
Thus:
a = 500/(1.025^6) = 431.148433 = 431.15$
3. a hypothetical strain of bacteria doubles every 5 minutes (exponential growth). one single bacterium was put in a sealed bottle at 9 a.m. and the bottle was filled at exactly 10 a.m. at what time was the bottle one-half full? think in terms of doubling time.
Denote n as the number of bacteria.
Then:
n = 1*1.05^t, where t the number of times 5 minutes have passed.
Denote T as the time on which the bottle full.
Then:
n(T) = 1*1.05^T.
n(T½) = 1*1.05^(T½). T½ = something too tired sooryy:P
After one year, the amount of money in the bank is:
1500*1.05.
After two years:
1500*1.05*1.05 = 1500*1.05^2.
Generally:
1500*10.5^t, where t is the time that has passed in years.
2. if i have $500 in my account after 4 years of investing at 2.5% per year, how much money did i start with?
using the same form as above:
a*1.025^t = 500.
a = 500/(1.025^t), where t=6 and a is the starting amount.
Thus:
a = 500/(1.025^6) = 431.148433 = 431.15$
3. a hypothetical strain of bacteria doubles every 5 minutes (exponential growth). one single bacterium was put in a sealed bottle at 9 a.m. and the bottle was filled at exactly 10 a.m. at what time was the bottle one-half full? think in terms of doubling time.
Denote n as the number of bacteria.
Then:
n = 1*1.05^t, where t the number of times 5 minutes have passed.
Denote T as the time on which the bottle full.
Then:
n(T) = 1*1.05^T.
n(T½) = 1*1.05^(T½). T½ = something too tired sooryy:P
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1. Year 1: $1500 * 5% = $75 interest. $1.500 + $75 = $1,575 at the end of year 1
Year 2: $1,575 * 5% = $78.75 interest. $1,575 + $78.75 = $1,653.75
Year 3: $1653.75 * 5%....just continue these steps for 6 years
2. Same as above, but working backwards. Each year you started with only 97.5% of today's total (100% - 2.5% = 97.5%)
Year 4: $500 * 97.5% = $487.50
Year 3: $487.50 * 97.5% = $475.31
Year 2: $475.31 * 97.5%...just continue until you have subtracted the amount earned in Year 1.
Year 2: $1,575 * 5% = $78.75 interest. $1,575 + $78.75 = $1,653.75
Year 3: $1653.75 * 5%....just continue these steps for 6 years
2. Same as above, but working backwards. Each year you started with only 97.5% of today's total (100% - 2.5% = 97.5%)
Year 4: $500 * 97.5% = $487.50
Year 3: $487.50 * 97.5% = $475.31
Year 2: $475.31 * 97.5%...just continue until you have subtracted the amount earned in Year 1.