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:
2.30/7.5
Step-by-step explanation:
4.30-2=2:30
2.5-10=7.5
Answer:
7b+5
Step-by-step explanation:
10b + 7 − 3b −2
=10b + 7 + −3b + −2
Combine Like Terms:
=10b + <u>7 </u>+ −3b + <u>−2
</u>
=(10b + −3b) + ( <u>7 </u>+ <u>−2</u>)
=7b + <u>5</u>
Answer:
7b + 5
Hope this helps : )
