For an F-distribution, find (a) f0.05 with v1 = 7 and v2 = 15; (b) f0.05 with v1 = 15 and v2 = 7: (c) f0.01 with v1 = 24 and v2 = 19; (d) f0.95 with v1 = 19 and v2 = 24; (e) f0.99 with v1 = 28 and v2 = 12.

In this exercise we have to calculate the distribution of F , so for certain values we will have:

A.)

B.)

C.)

D.)

E.)  

Using the F - distribution table find :

(a) For F equal to 0.05, we have:

b)For F equal to 0.05, we have:

(c)For F equal to 0.01, we have:

(d)For F equal to 0.95, we have:

(e)For F equal to 0.99, we have:

See more about distribution at link

A.) 2.71

B.) 3.51

C.) 2.92

D.) 0.47

E.) 0.35

Step-by-step explanation:

Using the F - distribution table :

find :

(a) f0.05 with v1 = 7 and v2 = 15

α = 0.05 ; v1 = 7 ; v2 = 15

F0.05 = 2.7066

b) f0.05 with v1 = 15 and v2 = 7:

α = 0.05 ; v1 = 15 ; v2 = 7

F0.05 = 3.5107

(c) f0.01 with v1 = 24 and v2 = 19;

α = 0.01 ; v1 = 24 ; v2 = 19

F0.01 = 2.92

(d) f0.95 with v1 = 19 and v2 = 24;

F0.95, (v1 = 19, v2= 24) = 1/[(F(1-0.95), v1=19, v2= 24]

Hence,

1/F0.05, v1 = 19, v2 = 24

= 0.47 (F distribution calculator)

(e) f0.99 with v1 = 28 and v2 = 12

F0.99, (v1 = 28, v2= 12) = 1/[(F(1-0.99), v1=28, v2= 12]

Hence,

1/F0.01, v1 = 28, v2 = 12

= 0.35 (F distribution calculator)

$2,703.06

step-by-step explanation:

add: $2,687.44+ $2,613.09+$2,808.64= $8,109.17

divide by 3 because you added 3 numbers.

$8,109.17/3=

answer: $2,703.06

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

=====================================

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).