Which situation could be represented by the graph?

a. students in the children’s choir are at least 8 years old but no more than 11 years old.
b. babysitters earn between $8 and $11 per hour.
c. a program at a community college can be completed in no fewer than 8 months, but must be completed in less than 11 months.
d. the dogs at a veterinary clinic are all 8 years old or under or are older than 11 years.

The correct option which resembles with the given graph is .

Further explanation:

From the given graph the it is observed that the interval is which is a half open interval or it can be said that it is a half closed interval.

A half closed or a half open interval is an interval in which one end point of the interval is included but the other point is not included.

For example:

In an interval of the form the point is included in the set but the point is not included in the set.

Similarly, in the interval the point is included in the set but the point is not included in the set.

In order to determine the correct option which resembles with given graph form interval for each option.

Option A:

As per the statement in option A the age of the student must be at least years and not more than years that is the age of the student is either years or greater than years and less than years.

As per the above statement the interval formed is in which is included in the set and is not included which means it is a half closed or half opoen interval.

Therefore, is a correct option.

Option B:

As per the statement in option B the amount of money the babysitter can earn is between and per hour.

As per the above statement the interval formed is in which both of the end points are not included in the set which means that it is an open interval.

Therefore, is an incorrect option.

Option C:

As per the statement in option C the time required to complete a program in college cannot be less than hours and the time must be less than hours.

From the above statement it is concluded that the time required can be hours or more than hours but it has to be less than hours.

So, as per the above the statement the interval formed is .

Therefore, is a correct option.

Option D:

As per the statement in option D the age of the dogs is either years old or below years and age of the dog must be greater than   tears.

As per the above statement the interval formed is .

Therefore, the is an incorrect option.

So, are the correct options which resembles with the given graph.

Learn more:

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Sets

Keywords: Interval, set, open interval, closed interval, half open interval, half closed interval, set theory, program, community college, babysitters, veterinary clinic, ages, students.

b

step-by-step explanation:

the word superfluous means unnecessary

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).