Answer:
Step-by-step explanation:
Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as
z = (x - u)/s
Where
x = ticket sales from monday
u = mean amount of ticket
s = standard deviation
From the information given,
u = 500 tickets
s = 50 tickets
We want to find the probability that the mean will be greater than 510. It is expressed as
P(x greater than 510) = 1 - P(x lesser than or equal to 510)
For x = 510
z = (510 - 500)/50 = 0.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.9773
P(x greater than 510) = 1 - 0.9773 = 0.0227
98 days = (98 ⁄ 7) weeks = 14 weeks
<span>Po = initial population = 5 </span>
<span>Ƭ = doubling time in weeks </span>
<span>t = elapsed time in weeks </span>
<span>P{t} = population after "t" weeks </span>
<span> P{t} = (Po)•2^(t ⁄ Ƭ) </span>
<span> P{t} = (Po)•2^(t ⁄ 4) </span>
<span> P{t} = 5•2^(t ⁄ 4) </span>
<span> P{14} = (5)•2^(14 ⁄ 4) … t = 14 weeks = 98 days </span>
<span> P{14} = 56 … population after 14 weeks</span>
Answer
Step-by-step explanation:
Answer:
t≈17
Step-by-step explanation:
