Which situation could be modeled by a linear function?

1. The value of a car depreciates by 7% annually.
2. A gym charges a [tex]$50 initial fee and then $[/tex]30 monthly.
3. The number of bacteria in a lab doubles weekly.
4. The amount of money in a bank account increases by 0.1% monthly.



Answer :

To determine which situation could be modeled by a linear function, let's analyze the options one by one:

1. The value of a car depreciates by 7% annually:
- Depreciation by a percentage of the remaining value each year is exponential, not linear. Each year, the value decreases by multiplying the current value by 0.93 (since 100% - 7% = 93%), resulting in an exponential decay.

2. A gym charges a \[tex]$50 initial fee and then \$[/tex]30 monthly:
- This can be modeled linearly. The total cost ([tex]\(y\)[/tex]) as a function of months ([tex]\(x\)[/tex]) can be expressed as:
[tex]\[ y = 30x + 50 \][/tex]
Here, the \[tex]$50 is a one-time initial fee (the y-intercept) and the \$[/tex]30 is the monthly fee (the slope). This forms a straight line when plotted.

3. The number of bacteria in a lab doubles weekly:
- Doubling each week represents exponential growth. The number of bacteria grows by a factor of 2 every week, meaning it multiplies the current number by 2, leading to exponential growth.

4. The amount of money in a bank account increases by 0.1% monthly:
- This increase is also exponential. If the interest is compounded monthly, the account balance increases by multiplying the current balance by 1.001 (since 100% + 0.1% = 100.1%), which models exponential growth.

Given this analysis, the situation that could be modeled by a linear function is:
2. A gym charges a \[tex]$50 initial fee and then \$[/tex]30 monthly.