Mathematics, 02.10.2019 04:30 quintinjerome

Ms. leverenz is doing an art project with her class. she has a 3 foot piece of ribbon. if she gives each student an eighth of a foot of ribbon, will she have enough for her class of 22 students?

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Answers

Answer from: jadaroyval

Yes, she will.

Further explanation

Given:

Ms. Leverenz is doing an art project with her class.

She has a 3-foot piece of ribbon. She gives each student an eighth of a foot of ribbon.

Question:

Will she have enough for her class of 22 students?

The Process:

$\boxed{ \ An \ eighth \ is \ \frac{1}{8} \ or \ equal \ to \ \frac{1 \times 125}{8 \times 125} = \frac{125}{1,000} = 0.125 \ }$

Let us draw a diagram that connects between she gives each student an eighth of a foot of ribbon with 22 students in her class.

$\boxed{ \ \frac{1}{8} \ }\boxed{ \ \frac{1}{8} \ }\boxed{ \ \frac{1}{8} \ } \cdot \cdot \cdot \cdot \boxed{ \ \frac{1}{8} \ }\boxed{ \ \frac{1}{8} \ } \\\boxed{1^{st}} \boxed{2^{nd}} \boxed{3^{rd}} \cdot \cdot \cdot \cdot \boxed{21^{st}}\boxed{22^{nd}} \ student$

From the diagram above, let us calculate how long (in ft) the ribbon she has given to 22 students.

$\boxed{ \ \frac{1}{8} \times 22 = ? \ }$

$\boxed{ \ = \frac{22}{8} \ }$

$\boxed{ \ = \frac{16}{8} + \frac{7}{8} \ }$

$\boxed{ \ = 2 + \frac{7}{8} \ }$

$\boxed{\boxed{ \ = 2\frac{7}{8} \ }}$

Therefore, there are $2\frac{7}{8}$ -foot piece of ribbon that she has given.

Ms. Leverenz has a 3-foot piece of ribbon.

$\boxed{ \ 2\frac{7}{8} < 3 \ }$

Thus, because the total number of pieces of ribbon given is smaller than the initial length, she will have enough for her class of 22 students

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Answer from: gacyber

She has 36 inches ribbon and she needs 33 inches for 22 students so she have enough for her class.

Step-by-step explanation:

1 feet = 12 inches or 30.48 cm.

Ms. Leverenz has 3 foot piece of ribbon for an art project.

3 foot = 12 × 3 = 36 inches

she gives each student an eighth of a foot

That is 8/12 inches = 1.5 inches to each student.

Ms. Leverenz gives each student 1.5 inches of ribbon.

Therefore she needs for 22 students

= 22 × 1.5 = 33 inches

She has 3 foot ribbon or 36 inches, so she have enough for her class of 22 students.

Answer from: Quest

step-by-step explanation:

21-4

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ 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).