Aprincipal evenly distributes 6 reams of copy paper to 8 fifth-grade teachers.
a. how many reams of paper does each fifth-grade teacher receive? explain how you know using pictures, words, or numbers.
b. if there were twice as many reams of paper and half as many teachers, how would the amount each teacher receives change? explain how you know using pictures, words, or numbers.

We start with 6 reams of copy paper and 8 teachers.

a) if we distribute the paper equally between all the teachers, then each teacher receive 6/8  (here we are dividing the total number of reams by the number of people that we are are splitting it). but 6/8 is less than one, so we must open the reams in order to split them.

b) If there were twice as many reams of paper, then we will have 2*6 = 12 reams of paper, and if there where half the number of teachers, then we have 8/2 = 4 teachers.

Doing the same procedure as before, we have 12 teams for 4 teachers, then each teacher has 12/4 = 3 reams

before, each teacher received 6/8 reams, and now, each one receive 3.

if we want to know the change, we need to see the difference between these numbers, so 3 - 6/8 =  3*8/8 - 6/8

notice that i am multiplying the 3 by 8/8, this is one, but  this way i can have the same denominator in both numbers.

24/8 - 6/8 = 18/8 = 9/4 = 8/4 + 1/4 = 2 + 1/4

so each teacher receive 2 + 1/4 reams than before.

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%

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).