General PostsFilesClass Notebook Assignments GradesFileHomeInsertDrawViewHelpOpen in Browser vO Tell me what you want to deI UCalibri11...State whether each situation is best represented by an exponentialor linear function. Then write an exponential or linear function forthe model and state whether the model is increasing or decreasing.1.Enrollment at a school is initially 454 students and grows by 3% per year.A salesperson initially earns $50,434 dollars per year and receives a yearly raiseof $675.2.3.A customer borrows $450 at 5% interest compounded annually.4.A wildlife park has 35 zebras and sends 1 zebra to another wildlife park eachyear.5. The value of a house is $546,768 and decreases by 3% each year.6.The population of a town is 66,666 people and decreases by 160 people each year.7.A business has a total income of $236,000 and revenues go up by 6.4% per year.

Answered: State whether each situation is best…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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O Tell me what you want to de
I U
Calibri
11
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State whether each situation is best represented by an exponential
or linear function. Then write an exponential or linear function for
the model and state whether the model is increasing or decreasing.
1.
Enrollment at a school is initially 454 students and grows by 3% per year.
A salesperson initially earns $50,434 dollars per year and receives a yearly raise
of $675.
2.
3.
A customer borrows $450 at 5% interest compounded annually.
4.
A wildlife park has 35 zebras and sends 1 zebra to another wildlife park each
year.
5. The value of a house is $546,768 and decreases by 3% each year.
6.
The population of a town is 66,666 people and decreases by 160 people each year.
7.
A business has a total income of $236,000 and revenues go up by 6.4% per year.

Expert Solution

Step 1

(1) Enrollment at school is initially 454 students and grows by $3 %$ per year.

Formula:

Next value $= Current value + \frac{Current value \times 3}{100}$

The data table is:

$x$ (year)	$y$ (students)
0	454
1	467.62
2	481.6486
3	496.098058
4	510.98099974
5	526.310429732

Since, $467.62 - 454 = 13.62$ , $481.6486 - 467.62 = 14.0286$

(a) It is clear that the change is not linear. Therefore, this situation represents an exponential function.

(b) We know that an exponential function is given by $y = a r^{x}$ , where $a = initial value, r = commom ratio$

Here, $a = 454$ and

$\frac{467.62}{454} = 1.03, \frac{481.6486}{467.62} = 1.03, \frac{496.098058}{481.6486} = 1.03, . . .$

Therefore, $r = 1.03$

The equation for the model is

$y = 454 {(1.03)}^{x}$

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