Answer:
1/hour =9$
For 12 hours she makes 108$
Answer:
P = 6200 / (1 + 5.2e^(0.0013t))
increases the fastest
Step-by-step explanation:
dP/dt = 0.0013 P (1 − P/6200)
Separate the variables.
dP / [P (1 − P/6200)] = 0.0013 dt
Multiply the left side by 6200 / 6200.
6200 dP / [P (6200 − P)] = 0.0013 dt
Factor P from the denominator.
6200 dP / [P² (6200/P − 1)] = 0.0013 dt
(6200/P²) dP / (6200/P − 1) = 0.0013 dt
Integrate.
ln│6200/P − 1│= 0.0013t + C
Solve for P.
6200/P − 1 = Ce^(0.0013t)
6200/P = 1 + Ce^(0.0013t)
P = 6200 / (1 + Ce^(0.0013t))
At t = 0, P = 1000.
1000 = 6200 / (1 + C)
1 + C = 6.2
C = 5.2
P = 6200 / (1 + 5.2e^(0.0013t))
You need to change the exponent from negative to positive.
The inflection points are where the population increases the fastest.
A. The expression given in the question is
c = 6x + (900000/x)
c = (6x^ +900000)/x
I hope that this is the expression that you were looking for.
b. When x = 240, then
c = [6 * (240)^2 + 900000]/ 240
= (6 * 57600) + 900000/240
= (345600 + 900000)/240
= 5190
From the above deduction, it can be concluded that the cost of ordering and storing is 5190. I hope the procedure is clear enough for you to understand.
Answer:
D
Step-by-step explanation:
Plz mark brainliest
