The yearly attendance at a local restaurant is 48,300 and grows continuously at a rate of 6.8% each year. What is the approximate attendance at the restaurant in 18 years?
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Seems like it's a problem of exponential growth.
To solve this you must first take the original amount of customers/year (so 48,300). Then you take 6.8%, which is the percentage of increase per year. So that means after 1 year, you would have 6.8% more customers than the year before, you have 1.068 times the original amount of customers.
So for one year it would be 48,300 * 1.068
But you want 18 years, so you simply write 48,300 * 1.068^18 (=157,844.58)
The basic formula for these types of taskes is N(t) = N(0) * a^t with a=e^k
Hope I could help,
~ Nick
To solve this you must first take the original amount of customers/year (so 48,300). Then you take 6.8%, which is the percentage of increase per year. So that means after 1 year, you would have 6.8% more customers than the year before, you have 1.068 times the original amount of customers.
So for one year it would be 48,300 * 1.068
But you want 18 years, so you simply write 48,300 * 1.068^18 (=157,844.58)
The basic formula for these types of taskes is N(t) = N(0) * a^t with a=e^k
Hope I could help,
~ Nick
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x = 48,300 * (1.068)^18
x = 48,300 * 3.268003767
x = 157,845
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x = 48,300 * 3.268003767
x = 157,845
=======================================…
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=48,300*(1.068)^18
=157,844
=157,844