Question

Devise the exponential growth function that fits the given​ data, then answer the accompanying question. Be sure to identify the reference point ​(t= ​0) and units of time. Between 2005 and 2010 ​, the average rate of inflation in a certain country was about 4 ​% per year. If a cart of groceries cost ​$190 in 2005 ​, what will it cost in 2015 assuming the rate of inflation remains​ constant?What is the reference point (t=0)? a. The inflation rate, 5% b. The cost of groceries $120. c. The year 2010. d. The year 2005. e. The year 2015.

Answer 1

Answer:

The cost in 2015 will be $283.45.

What is the reference point (t=0)?

d. The year 2005.

Step-by-step explanation:

The exponential growth model for the cost is the following:

$C(t) = C_{0}e^{rt}$

In which C(t) is the cost after t years, $C_{0}$ is the initial cost and r is the decimal inflation rate

In this problem, we have that:

$C_{0} = 190, r = 0.04$

What will it cost in 2015 assuming the rate of inflation remains​ constant?

2015 is 10 years after 2005, so this is C(10).

$C(t) = C_{0}e^{rt}$

$C(10) = 190*e^{0.04*10} = 283.45$

The cost in 2015 will be $283.45.

The reference point is the year in which t = 0, so the year 2005.

Answer 2

Answer:

Y(t) = 190*e^{5*0,04}

Step-by-step explanation:

Exponential equations, are very helpfull tool when dealing with growing or decaying process. The most common shape of such equatio is:

Y(t)  =  Y₀*e^{kt}    

In that equation:

Y(t) would be the value of Y ( whatever the situation is ) at certain time t

Y₀  stand for value of Y at the time t =t₀ = start of the evaluating process

e is the well known number =  2.718

k is a parameter (indicator of the number of units in which the rate of growing is measured ( in our case the rate of growing in % is 4 we have to express it as a number 0,04  and the number of years is 5 from 2005 up to 2010, taken into account the day and the month is the same for the two measured)

t is the rate of growing or decaying of the process.

In our particular case we have

Y(t) would be the cost of a cart of groceries at time t

Y₀ is the value of Y when  t = 0   one day in 2005

Hear we have to be aware of the fact that problem statement does not identify exact date but only years , and the rate of inflation is given per year. So we have to assume that dates are over the same day and month in order to have a complete year for the rate of growing.

If we do so (for instance from 12-06-2005 and 12-06-2010) we get 5 complete years and are able to say that we are dealing with complete 5 years.

Then our final equation is

Y(t) = Y*e^kt}      Y(t) = 190*e^{5*0,04}