Mathematics, 01.08.2019 23:00, youngg52

Find the bank discount and proceeds using ordinary interest on an unsecured promissory note made to joan williams for $10,000 at 5.5% annual simple interest from july 28 september 10 for this year. use the steps below to find your answer

Answers: 1

Show answers

Answers

If these were the missing questions:

a)Explain the difference between a simple interest note and a simple discount note.
We can identify the difference between these two by using formula:
Simple Interest Note:
Interest = Face value or Principal * Rate * Time
Maturity Value = Principal + Interest
Proceeds = Principal
Effective Rate = Rate / (Proceeds * time)

Simple Discount Note:
Interest = Face Value or Principal * Rate * Time
Maturity Value = Face Value
Proceeds = Maturity Value - Bank Discounts
Effective Rate = Rate / (Proceeds * time)

In simple interest note, the borrower will have to pay for the principal plus the interest. In simple discount note, since the proceeds has already been net of discount, the borrower will pay for the face value of the note.

b)Exact time (days) of note:
July has 31 days. July 28 is the start of the loan.
August has 31 days.
September has 30 days. September 10 is the payment of the loan.

31-27 = 4 days for July
4 days for July + 31 days for August + 10 days for September = 45-day term of the note.

c)What is the bank discount?
Interest = Principal * Rate * term
Interest = 10,000 * 0.055 * 45/360 = 68.75

The bank discount is 68.75

d)What are the proceeds Joan receives?
If Simple Interest Note, Joan receives $10,000.
If Simple Discount Note, Joan receives $9,931.25 → (10,000 - 68.75)

e)What is the amount Joan repays?
If Simple Interest Note, Joan will pay $10,068.75
If Simple Discount Note, Joan will pay $10,000

answered: Guest

40/5 = 8you just divide the number of gloves by the number of teams

answered: Guest

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).