There are 1800 students in a school this number is 20% more than what they had last year find the number of students in the scho
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A dot above 7, 4, 10, 2, and 20. Draw a box around the median, first quartile, and and the third quartile. Then connect the box to the minimum and maximum by drawing a line. Should look like this:
Answer:
(a)
(b) (i) The population after 8 hours is 607,325
(ii) The population after 24 hours is 1,828,643
(c) The rate of increase of the population as a percentage per hour is 7.132%
(d) The doubling time of the population is approximately, 10.06 hours
Step-by-step explanation:
(a) The initial population of the bacteria, y₁ = a = 350,000
The time the colony grows, t = 12 hours
The final population of bacteria in the colony, y₂ = 800,000
The exponential growth model, can be written as follows;
Plugging in the values, we get;
Therefore;
(1 + r)¹² = 800,000/350,000 = 16/7
12·㏑(1 + r) = ㏑(16/7)
㏑(1 + r) = (㏑(16/7))/12
r = e^((㏑(16/7))/12) - 1 ≈ 0.07132
The model is therefore;
(b) (i) The population after 8 hours is given as follows;
y = 350,000 × (1 + 0.07132)⁸ ≈ 607,325.82
By rounding down, we have;
The population after 8 hours, y = 607,325
(ii) The population after 24 hours is given as follows;
y = 350,000 × (1 + 0.07132)²⁴ ≈ 1,828,643.92571
By rounding down, we have;
The population after 24 hours, y = 1,828,643
(c) The rate of increase of the population as a percentage per hour = r × 100
∴ The rate of increase of the population as a percentage = 0.07132 × 100 = 7.132%
(d) The doubling time of the population is the time it takes the population to double, which is given as follows;
Initial population = y
Final population = 2·y
The doubling time of the population is therefore;
Therefore, we have;
2·y/y =2 =
t = ln2/(ln(1 + 0.07132)) ≈ 10.06
The doubling time of the population is approximately, 10.06 hours.