The data set monarch from computer-active data analysis by lunn andmcneil (1991) contains the years lived after inauguration, election, or coronation of u. s. presidents, popes, and britishmonarchs from 1690 to roughly the present (nixon, carter, reagan, etc., are not included). do the groups differ? thestats output is
> > stats(monarch. years, monarch. group)

pres popes k& qs
n 35. 23. 14.
mean 14.5429 12.5652 22.7143
std. dev. 9.4662 8.7221 18.6028
suggesting that the monarchs tend to live longer than presidents orpopes. this is not experimental data (but instead observational)and one could claim that the data is nearly the population. nevertheless, we have run the anova and the results look like
source df ss ms f p-val

group 2 x x 3.5880 ?
error 69 x x
using the fact that this f statistic has an f density with 2 and 69degrees of freedom under random sampling and the null hypothesis ofequal means, find the p value using splus or matlab and the factthat pr(f > y) = 1-pf(y, df1,df2) in splus and pr(f > y) =1-fcdf(y, df1,df2) in matlab, where df1 is the numerator degrees offreedom and df2 is the denominator degrees of freedom in the fstatistic. p value =

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).

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8

it is not that hard all u have to do is plot the graph

step-by-step explanation: