Solve and round to the nearest hundredth 24.5 = - 2.5c - 4c
1 answer:
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Draw a diagram to illustrate the problem as shown in the figure below.
The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.
The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
The mean depth is k = (2.5 + 5.5)/2 = 4.0 m
The model for tide depth is
That is,
d = -1.5 cos(0.5027t) + 4
where
d = depth, m
t = time, hours
A plot of the function confirms that the model is correct.
The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.
The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
The mean depth is k = (2.5 + 5.5)/2 = 4.0 m
The model for tide depth is
That is,
d = -1.5 cos(0.5027t) + 4
where
d = depth, m
t = time, hours
A plot of the function confirms that the model is correct.
Answer:
The initial population was 2810
The bacterial population after 5 hours will be 92335548
Step-by-step explanation:
The bacterial population growth formula is:
where P is the population after time t, is the starting population, i.e. when t = 0, r is the rate of growth in % and t is time in hours
Data: The doubling period of a bacterial population is 20 minutes (1/3 hour). Replacing this information in the formula we get:
Data: At time t = 100 minutes (5/3 hours), the bacterial population was 90000. Replacing this information in the formula we get:
Data: the initial population got above and t = 5 hours. Replacing this information in the formula we get: