Time answer 15 points

jolianne worked half as many hours as andrew. jolianne makes $10 per hour and has saved $27. andrew makes $8 per hour and has no money saved. after getting paid, they have the same amount of money.

andrew claims that only his table and equation are correct, but jolianne claims that both tables and equations are valid. which statement regarding their claims is true?

andrew is correct because t cannot be used to represent the number of hours jolianne worked.

andrew is correct because jolianne worked half as many hours as he did and he did not work twice as many hours as she did.

jolianne is correct because if one lets t = hours andrew worked, then his equation is valid. if one lets t = hours jolianne worked, then andrew worked twice as many so her equation is valid, too.

jolianne is correct because when solving t in her equation, t = hours jolianne worked, and when solving for t in andrew’s equation, t = hours jolianne worked so they are equal.

Andrew is correct because Jolianne worked half as many hours as he did and he did not work twice as many hours as she did

Step-by-step explanation:

Let

x ----> number of hours worked by Jolianne

t ----> number of hours worked by Andrew

we know that

The number of hours worked by Jolianne multiplied by $10 per hour plus $27 saved must be equal to the number of hours worked by Andrew multiplied by $8

The linear equation that represent this situation is

----> equation A

-----> equation B

substitute equation B in equation A

so

Andrew's table and equation is correct

Jolianne's table and equation are not correct, because Andrew did not work twice as many hours as she did

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).