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eimsori [14]
2 years ago
12

Edgar accumulated $5,000 in credit card debt. If the interest rate is 10% per year and he does not make any payments for 3 years

, how much will he owe on this debt in 3 years by compounding continuously?
Mathematics
2 answers:
victus00 [196]2 years ago
7 0

the discrete compounding formula is f = p * (1 + r) ^ n

f is the future value

p is the present value

r is the interest rate per time period

n is then number of time periods

in your problem, you are given:

f = what you want to find

p = 5000

r = 30% per year / 100 = .3 per year (percent / 100 = rate).

n = 3 years

if you compound annually, the formula becomes:

f = 5000 * (1 + .3) ^ 3 = 10985

if you compound quarterly, the formula becomes:

f = 5000 * (1 + .3 / 4) ^ (3 * 4) = 11908.898

if you compound monthly, the formula becomes:

f = 5000 * (1 + .3 / 12) ^ (3 * 12) = 12162.67658

if you compound continuously, a different formula is used.

that formula is f = p * e ^ (r * n)

f is the future value

p is the present value

e is the scientific constant of 2.718281828.......

r is the interest rate per time period

n is the number of time periods.

with this formula, you leave the time periods in terms of years.

it will make no difference what time periods and compounding periods you use, the answer will be the same.

most of the time you will just give it the interest rate per year and the number of years.

the reason is as follows:

r * n = .3 * 3 = .9 when giving it rate and time in terms of years.

r * n = .3 / 4 * 3 * 4 = .9 when giving it rate and time in terms of quarters.

r * n = .3 / 12 * 3 * 12 = .9 when giving it rate and time in terms of months.

in your problem, the formula becomes f = 5000 * e ^ (.3 * 3) = 12298.01556.

the more compounding periods per year, the higher the future value.

the highest is when you compound continuously.

this is apparent from the data.

amm18122 years ago
5 0
Yyyy please please thank you lord lord thank
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5 0
3 years ago
Bacteria of species A and species B are kept in a single environment, where they are fed two nutrients. Each day the environment
DiKsa [7]

Answer:

We require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

Step-by-step explanation:

Let n₁ be the population of A required and n₂ be the population of B required.

Now we require 2 units of the first nutrient for species A and one unit of the first nutrient for species B. The total nutrients required by species A is 2n₁ and that by species B is 1n₂ = n₂. So, the total nutrients required by both species A and B is 2n₁ + n₂. Since this equals the quantity of the first nutrient which is 10,560, then  2n₁ + n₂ = 10,560 (1)

Now we require 5 units of the second nutrient for species A and 6 units of the second nutrient for species B. The total nutrients required by species A is 5n₁ and that by species B is 6n₂. So, the total nutrients required by both species A and B is 5n₁ + 6n₂. Since this equals the quantity of the first nutrient which is 31,510, then  5n₁ + 6n₂ = 31,510 (2).

So, we have two simultaneous equations which we would solve to find the populations of A and B which satisfy both equations.

2n₁ + n₂ = 10,560  (1)

5n₁ + 6n₂ = 31,510 (2)

From (1) n₂ = 10,560 - 2n₁ (3)

Substituting equation (3) into (2), we have

5n₁ + 6(10,560 - 2n₁) = 31,510

expanding the brackets, we have

5n₁ + 63,360 - 12n₁ = 31,510

collecting like terms, we have

5n₁ - 12n₁ = 31,510 - 63,360

simplifying, we have

- 7n₁ = -31,850

dividing both sides by -7, we have

n₁ = -31,850/-7

n₁ = 4,550

Substituting n₁ = 4,550 into (3), we have

n₂ = 10,560 - 2(4,550)

n₂ = 10,560 - 9,100

n₂ = 1,460

So, we require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

3 0
2 years ago
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