1. t =23 will give you 8230
2. since it is decaying the equation is P=Ae^-kt
initially A = 1 but it decays to 0.25 after 68 min
thus k =-1/68ln(0.25)
when t =180
P=e^1/68ln(0.25)x180 =0.0254x100 = 2.5%
1) The population P of a town rose from 1000 at the beginning of 1975 to 2500 at the beginning of 1985. The natural growth is given by P = 1000e^0.1 ln 2.5t.
a) What was the population of the town at the beginning of 1998, correct to the nearest 10 people?
(I worked it out to be 8300 but the solutions says itâ€™s 8230)???
b) find the rate dP/dt at which the population is increasing at the beginning of 2000 correct to the nearest whole number?
( so I worked out dP/dt = 0.1ln 2.5x1000e^0.1ln2.5x25 so when t=25 dP/dt = 918 but the solution says 916??)
2) A certain radioactive isotope decays at such a rate that after 68 mins only a quarter of the initial amount remains. What proportion of the initial amount will remain after 3 hours? Give your answer as a percentage.
3) Given that C = 20 000 e^1/5ln9/8x. Where x is the distance in metres and C is concentration of gas in parts per million (ppm). The accepted safe level for this gas is 30 ppm. How far should everyone be from the gas leak. Rounding to nearest 10 m.
( I got 271m but the solution says 330m)??? How???
1. t =23 will give you 8230
2. since it is decaying the equation is P=Ae^-kt
initially A = 1 but it decays to 0.25 after 68 min
thus k =-1/68ln(0.25)
when t =180
P=e^1/68ln(0.25)x180 =0.0254x100 = 2.5%
Thank you what about 1b I get 905 í ½í¸©
i got 905 as well, perhaps the answer is wrong
HSC 2018
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