Ms. Haven recorded the number of homework assignments her students completed and their test grades. Her findings are shown in the table. Which ordered pair is an
outlier in this data set?

(6, 97) is the outlier.

Step-by-step explanation:

When we have a data set:

x: x₁, x₂, x₃, ..., xₙ

y: y₁, y₂, y₃, ..., yₙ

Each one of the pairs (x₁, y₁), (x₂, y₂), etc, is an ordered pair that belongs to the data set.

Usually, these values are distributed in some way that, we assume, is normalized. So if one of these ordered pairs is really different from the others, then that pair is called an "outlier".

In the data set, we have:

Number of homework completed vs Score assigned.

You can see that, as the number of homework completed increases, also does the score

For example, for the smaller number of homework completed, (5) the score is 67, the minimum score.

And as the number of homework completed reaches to 10, the score increases to 90 or 98.

But there is an outlier there, you can see that a student that only completed 6 times the homework, has a score of 97 (one of the largest scores) (this score is more than all the scores for the students with 6, 7, 8 and 9 completed homeworks) so we can conclude that this one is the outlier.

(6, 97) is the outlier.

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)

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

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).