Given the following experiment: Male Albino laboratory rats are used routinely in many kinds of experiments. Proper nutrition for the rats is important. Specifically, this experiment examines the amount of protein in the diet and the amount of threonine in the diet. The general protein in the diet is threonine deficient. There are three levels of threonine (0.3, 0.6, 0.9% of diet) and three levels of protein (12, 17, and 21% of diet), for a total of 9 treatments. Gender was a potentially significant factor but the researcher was not interested to compare between males and females. She considered 180 male rats. Those rates were all weighed and separated into five groups of 36 of approximately uniform weight (Weight at time zero). The rats in each group were randomly assigned to each of the nine treatments (one treatment per group). Rats’ body weights were measured three more times: after one, two and three months.
A) Answer the following:
i) What is the response variable?
ii) List the factors in this study and their levels
iii) What factor(s) was used as a blocking factor?
iv) What nuisance factor(s) that was held constant?
B) How many observations in this experiment? Since the error is a function of subjects (rats): Error (factor (Subject)). What is the degree of freedom for the residuals?
C) What kind of experimental design used in this experiment? List the between block verses within block nuisance factors for this experiment.
D) Draw the experiment design starting with the benchmark. Label each part along with its degrees of freedom.

ab will be equal to

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).

12

step-by-step explanation:

a triangle's   full angle is 180 degrees so 180 - 156 = 24. 24/2=12

a)(x-2)^2+(y-1)^2=9

step-by-step explanation:

(picture explains)