Answer:
<u><em>6565 = 1010(4) + 2525</em></u>
Step-by-step explanation:
so the equation would be
6565 = 1010x + 2525
x = 4 because
1010 (4) = 4040
and 4040 + 2525 = 6565, so
<u>6565 = 1010(4) + 2525</u>
Answer:
7,228,202%
Step-by-step explanation:
Population at first census in 1979 was 3929
If the Population had increased to 284,000,000 by 2000, the population increase will be;
284,000,000-3929
= 283,996,071
%increase in population = increment/initial population×100%
= 283,996,071/3929×100%
= 7,228,202%
This means that the population has increased by 7,228,202%
Answer:
3960.4 bacteria
Step-by-step explanation:
The formula to solve the above question is given as:
P(t) = Po (2) ^t/k
P(t) = Population after time t = ?
Po = Initial population = 650 bacteria
t = Time in days = 7.3 days
k = doubling time = 2.8 days
P(t) = 650 × (2)^7.3/2.8
P(t) = 650 × 2^2.6071428571
P(t) = 650 × 6.0929582599
P(t) = 3960.4228689 bacteria.
Approximately = 3960.4 bacteria
Therefore, the number of bacteria the researcher will have after 7.3 days if they started with 650 bacteria is 3960.4 bacteria.
Answer:
i think G
Step-by-step explanation:
This question is based on the concept of
amount of interest. Therefore, 22,234 would be
the population after 11 years, to the nearest
whole number.
Given:
A town has a population of 13000 and grows at
5% every year.
We need to determined the population after 11
years, to the nearest whole number.
We know that,
Formula =A = P(1+ r)
Where, p be the population. r is the rate and t is
the time.
P= 13000,
r= 5%
†= 11
Now put all these values in formula.
We get,
5
A = 13000(1 +
A
= 13000(1 + 0.05)11
A
= 1300022, 234.41(0.95)