Answer:
The number of bacteria at initial = 187
Step-by-step explanation:
Given that the population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t.
Integrating both side we get
㏑ N = k t + C ------- (1)
Now given that after 3 hours it is observed that 500 bacteria are present and after 10 hours 5000 bacteria are present.
⇒ ㏑ 500 = 3 k + C -------- (2)
⇒ ㏑ 5000 = 10 k + C ------ (3)
⇒ ㏑ 5000 - ㏑ 500 = 7 k
⇒ ㏑
= 7 k
⇒ ㏑ 10 = 7 k
⇒ k = 0.329
Put this value of k in equation (2),
⇒ ㏑ 500 = 3 × 0.329 + C
⇒ C = 5.23
Put this value of C in equation 1 we get,
⇒ ㏑ N = k t + 5.23
Initially when t = 0 , then
⇒ ㏑ N = 5.23
⇒ N = 187
Thus the number of bacteria at initial = 187
Answer: see below
Step-by-step explanation:
(i)
200/2 = 100
Find 100 on cumulative frequency and the correspinding time: 14 minutes. So the answer is 14 minutes.
(ii)
First, find the lower quartile —> 200/4 = 50
Upper quartile —> 200/(3/4) = 150
Find the difference: 150 - 50 = 100
So the interquartile range is 100
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Answer:
2x3=6+9=15?
Step-by-step explanation:
Im sorry if i didnt help.
It is less than 324.21, because as you can see, all the numbers are equal until you get to the 0 & 1, and 1 is greater than 0
Answer:
C, none of them are equivalent
BECAUSE
For A, the -4q should be positive 4q to apply
For B, the 2q when distributing with the parentheses should be negative 2q to work with the expression above
Hope this helps!
