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
Answer:
A
Step-by-step explanation:
Answer:
50%
Step-by-step explanation:
So there are 90 2 digit numbers. There are 45 2 digit odd numbers.
The probability should be 50%
Using the law os cosines formula b^2 = a^2 + c^2 - 2*a*c*cos(B)
a = 17, b = 8, c = 16
8^2 = 17^2 + 16^2 - 2*17*16* cos(B)
64 = 289 + 256 - 544 * cos(B)
544*cos(B) = 289 + 256 - 64
544 * cos(B) = 481
cos (B) = 481/544
B = arccos(481/544)
B = 27.8 degrees
