Mathematics, 18.07.2019 09:50 tddreviews

Tina wants to save money for school. tina invests $1,000 in an account that pays an interest rate of 6.75%. how many years will it take for the account to reach $18,600?

Answers: 1

Show answers

Answers

Assuming that the interest is paid as compound interest.

Final value = Intial Value × (Multiplier)ⁿ

Multiplier is 100% + the percentage increase/decrease
The letter 'n' is for the number of years

We have:
Initial Value = 1000
Final Value = 18600
Multiplier = 100% + 6.75% = 1 + 0.0675 = 1.0675

We are looking to find:
The number of years 'n'

Substituting the values we have into the formula

18600 = 1000 × (1.0675)ⁿ
18600 ÷ 1000 = 1.0675ⁿ
18.6 = 1.0675ⁿ ⇒ From here you can use either trial and error method or you can use logarithm law

Using trial and method
Let n = 1; 1.0675¹ = 1.0675   ⇒ Not the result we want
Let n = 5; 1.0675⁵ = 1.39   ⇒ Not the result we want
Let n = 10; 1.0675¹⁰ = 1.92 ⇒ Not the result we want
Let n = 25; 1.0675²⁵ = 5.12   ⇒ Still too small
Let n = 55; 1.0675⁵⁵ = 36.33   ⇒ Values are too big
Let n = 40; 1.0675⁴⁰ = 13.64   ⇒ Too small (we narrowed down the value of n to be between 40 and 55)

Let n = 48; 1.0675⁴⁸ = 23.00   ⇒ Too big
Let n = 45; 1.0675⁴⁵ = 19.00 ⇒ Let's try n = 44 to see which value are closer to 18.60

Let n = 44; 1.0675⁴⁴ = 17.70   ⇒ This a little far from 18.6

So we have our answer, n = 45

Conclusion: Tina needs to save for 45 years to get $1000 to $18600 with compound interest 6.75% paid yearly.

Answer from: Quest

well my understanding of this question is they want you to use the y=14.1x -2.3 and put the coordinates they gave you onto the line segment of y=14.1x -2.3

step-by-step explanation:

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).