Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
Step-by-step explanation:
It often works well to write an exponential expression as ...
value = (initial value)×(growth factor)^(t/(growth period))
(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...
P(t) = 100·2.3^t
__
(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours
__
(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t
P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour
__
(d) We want to find t such that ...
P(t) = 10000
100·2.3^t = 10000 . . . substitute for P(t)
2.3^t = 100 . . . . . . . . divide by 100
t·log(2.3) = log(100)
t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000
what are the measurments of the cone, but the formula to the cone is V
v = pi times radus times radius h/3
Most likely B because one pound is equal to 6 dollars. So i think it would be x= 1 and y= 6
So put the x as the dom and 6 as the num?
The maximum possible profit = $7068
For given question,
One Microsoft July $72 put contract for a premium of $1.32
The payoff arise from put option is max (K - S, 0) - P
Now it would be maximum at S = 0
And, the maximum payoff is
K - 0 - P
= K - P
= 72 - 1.32
= $70.68
We assume that for each and every contract the number of shares is 100
So, the maximum profit gained from this strategy is
= $70.68 × 100 shares
= $7068
The maximum profit that will be gained from this strategy is $7068
Therefore, the maximum possible profit = $7068
Learn more about the profit here:
brainly.com/question/20165321
#SPJ4
1) 3x=(4x+50-30)/2
x=10
Explanation: (1/2)(big arc - small arc), plug in the numbers and solve
2) 15•x=33•10
x=22
Explanation: a•b=c•d, plug in the numbers and solve
3) (x+8)•8=(24+4)•4
x=6
Explanation: W•E=W•E, whole•external part= whole•external, plug in the numbers and solve
