Best ontime airlines: hawaiian airlines topped the list of the most punctual u. s. airlines for full-year 2014, according to data released by the bureau of transportation statistics. the on-time arrival performance was 91.9% another aviation data service believes that this percentage now is even higher. the team of researchers chose 130 randomly selected flights and find that 120 of the flights arrived on time. can a hypothesis test be used to determine whether the proportion of hawaiian airlines flights that arrive on-time is higher than 91.9%? group of answer choices yes, because the sample is random. thus, it is representative of the population of flights for the airlines. yes, because the sample is random and (130)(0.923) and (130)(0.077) are both at least 10. this means the normal model is a good fit for the sampling distribution. yes, because the sample is random and (130)(0.919) and (130)(0.081) are both at least 10. this means the normal model is a good fit for the sampling distribution.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

table 3 represents a direct variation function. : )

the total estimated nappy use.=10,500,000

step-by-step explanation:

given that on an average the average baby uses 3528 napkins.

in a particular year 2014, the number of births were 299697 in new south wales.

rounding off the number of births to the nearest hundred thousand we have

no of birtsh = 300000

also rounding off the average usage to nearst thousand we get

average use of a baby = 3500

hence total number of nappy estimated to be used = 3500x300000

= 10,500,000

answer is 10,500,000 nappies