Mathematics: A. in 2000, the population of a country was approximately 5.755.75 million and by 20282028 it is projected to grow to 88 million. use the exponential growth model upper a equals upper a 0 e superscript kta=a0ekt, in which t is the number of years after 2000 and upper a 0a0 is in millions, to find an exponential growth function that models the data. b.b. by which year will the population be 77 million?

A. in 2000, the population of a country was approximately

Ответ:

ashlon98

14.03.2022

A. in 2000, the population of a country was approximately 5.75 million and by 2028 it is projected to grow to 88 million. use the exponential growth model to find an exponential growth function that models the data.
B. By which year will the population be 77 million?

To solve this we are going to use the exponential growth model: $A=A_{0}e^{kt}$
where

is the final population after

years
$A_{0}$ is the initial population

is the Euler's constant

is the growth rate

is the time in years

A. We know for our problem that A=88

, $A_{0}=5.75$ , and t=2028-2000=28

. So lets replace those values in our exponential growth model to find

:
$A=A_{0}e^{kt}$
$88=5.75e^{28k}$
$e^{28k}= \frac{88}{5.75}$
$ln(e^{28k})=ln( \frac{88}{5.75} )$
$28k=ln( \frac{88}{5.75} )$
$k= \frac{ln( \frac{88}{5.75}) }{28}$
k=0.09743

Now that we have

, we can find an exponential growth function that models the data by replacing $A_{0}$ and

in the exponential growth model:
$A=5.75e^{0.09743t}$
We can conclude that the equation that models the data is: $A=5.75e^{0.09743t}$ .

B. We know for a our problem that A=77

and $A_{0}=5.75$ . We also know from our previous calculation that k=0.09743

. So lets replace those values in our equation to find

:
$A=5.75e^{0.09743t}$
$77=5.75e^{0.09743t}$
$e^{0.09743t}= \frac{77}{5.75}$
$ln(e^{0.09743t})=ln( \frac{77}{5.75})$
$0.09743t=ln( \frac{77}{5.75})$
$t= \frac{ln( \frac{77}{5.75}) }{0.09743}$
t=26.6

We can conclude that the population of the country will reach 77 million by July of 2026.

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Ответ:

studybuddy0203

05.12.2022

Mathematics

the fraction of the cars manufactured that were not boxcars is 12/103

Step-by-step explanation:

A company that manufactures cars for freight trains manufactured 309 cars in the month of August. 273 of the cars manufactured by the company were boxcars. It means that the number of manufactured cars that were not boxcars would be

309 - 273 = 36

Therefore, the fraction of the cars manufactured that were not boxcars would be

36/309 = 12/103

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