Answer:
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters
Step-by-step explanation:
<u>Step 1</u>:-
Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.
by Poisson random variable given that λ = 1.5 flaws/millimeter
Poisson distribution 
<u>Step 2:</u>-
The probability that no flaws occur in a certain portion of wire

On simplification we get
P(x=0) = 0.223 flaws occur / millimeters
<u>Conclusion</u>:-
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters
For this case we have a function of the form:

Where,
A: initial amount
b: growth rate (if b> 1)
n: time in hours
Substituting values we have:

We have then that the initial amount is:

If b = 1.85 then the growth percentage is:
Answer:
here were initially 20 bacteria.
The hourly percent growth rate of the bacteria would be 85%
A. 2x3=6 and 6x12=72 so 78 is greater than 36
6+72=79 which is less than 90
write it out in the order they give
Julie , Roger
James , Julie, Roger
David, James, Julie, Roger
Sarah, David, James, Julie, Roger
David is in 2nd place