Answer:
a) The continuous rate of growth of this bacterium population is 45%
b) Initial population of culture at t = 0 is 950 bacteria
c) number of bacterial culture contain at t = 5 is 9013 bacteria
Step-by-step explanation:
The number of bacteria in a culture is given by
n(t) = 950
a) rate of growth is
Bacteria growth model N(t) = no
Where r is the growth rate
hence r = 0.45
= 45%
b) Initial population of culture at t = 0
n(0) = 950
= 950
= 950 bacteria
c) number of bacterial culture contain at t = 5
n(5) = 950
= 950
= 9013.35
= 9013 bacteria
Answer:
2) 4/3
Step-by-step explanation:
(4^-3 * 3^4 * 4^2)/(3^5 * 4^-2)
4^-3 * 3^4 * 4^2 = 4^-1 * 3^4
(4^-1 * 3^4)/(4^-2 * 3^5)
(3^4)/(3^5) = 1/(3^5-4)
(4^(-1))/(4^-2 * 3^(5-4)
(4^-1)/(4^-2 * 3)
4^(-1 - (-2)) / (3)
-1 - (-2) = 1
4/3
Answer:
x >4 or x < -12/5
Step-by-step explanation:
An absolute value has 2 solutions, a positive and a negative. Remember to flip the inequality for the negative solution
|5x-4|>16
5x-4 > 16 5x-4 < -16
Add 4 to each side
5x-4 +4> 16+4 or 5x-4+4 < -16+4
5x > 20 or 5x < -12
Divide each side by 5
5x/5 > 20/5 or 5x/5 < -12/5
x >4 or x < -12/5
Answer:
(a) 
The expected number in the sample that treats hazardous waste on-site is 0.383.
(b) 
There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.
Step-by-step explanation:
Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.
N = 209
Only eight of these facilities treated hazardous waste on-site.
r = 8
a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?
n = 10
The expected number in the sample that treats hazardous waste on-site is given by
Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.
b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site
The probability is given by
For the given case, we have
N = 209
n = 10
r = 8
x = 4
Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.
The same way you divide normal numbers, 0.5/0.5 = 1
