Answer:
here i finished!
hope it helps yw!
Step-by-step explanation:
The doubling period of a bacterial population is 15 minutes.
At time t = 90 minutes, the bacterial population was 50000.
Round your answers to at least 1 decimal place.
:
We can use the formula:
A = Ao*2^(t/d); where:
A = amt after t time
Ao = initial amt (t=0)
t = time period in question
d = doubling time of substance
In our problem
d = 15 min
t = 90 min
A = 50000
What was the initial population at time t = 0
Ao * 2^(90/15) = 50000
Ao * 2^6 = 50000
We know 2^6 = 64
64(Ao) = 50000
Ao = 50000/64
Ao = 781.25 is the initial population
:
Find the size of the bacterial population after 4 hours
Change 4 hr to 240 min
A = 781.25 * 2^(240/15
A = 781.25 * 2^16
A= 781.25 * 65536
A = 51,199,218.75 after 4 hrs
2 x 2 x 2 x 3
Because you are multiplying two, four times
It is 13,14 because it goes up two then one one then two again which will then mess up the rotation
Option (A) : least: 10 hours; greatest: 14 hours
The function f(x) = sin x has all real numbers in its domain, but its range is
−1 ≤ sin x ≤ 1.
How to solve such range questions?
Such questions in which every term is in addition and its range is asked is simplest ones to solve if we know the range of each of term. This can be seen from this question
Given: d(t) = 2sin(xt) + 12
= −1 ≤ sin (xt) ≤ 1.
= −2≤ 2 sin (xt) ≤ 2.
= 10 ≤ 2sin (xt) + 12 ≤ 14
= 10 ≤d(t) ≤ 14
Thus least: 10 hours; greatest: 14 hours
Learn more about range of trigonometric ratios here :
brainly.com/question/14304883
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