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iren [92.7K]
3 years ago
15

Francesca obtains a loan of $18.965.00. The loan has a simple interest rate of 3%. How much interest will Francesca owe at the e

nd of the first year?
Mathematics
1 answer:
marin [14]3 years ago
6 0

Answer:

$568.95

Step-by-step explanation:

Given parameters:

Value of loan  = $18965

Interest rate  = 3%

Time  = 1year

Unknown:

Interest = ?

Solution:

The interest on a particular amount is given as:

        I  = \frac{PRT}{100}  

I is the interest

P is the principal

R is the rate

T is the time

       I  = \frac{18965 x 3 x 1}{100}   = $568.95

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laila [671]

The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

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c) When wil the population reach 750?

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Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

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