Respuesta :
First, according to the information, we build a linear model, and then, the model is used to find the information asked.
a) The linear model is:
[tex]V(t) = 224000 - 6800t[/tex]
b) The depreciated value of the boat after 10 years is of $156,000.
c) The depreciated value falls below $100,000 after 18.24 years.
d) The sketch is given at the end of the answer.
Item a:
- The initial value is of $224,000, thus, the y-intercept of the linear function is of [tex]b = 224000[/tex].
- The value decayed to $115,200 in 16 years, thus, the slope of the linear function is:
[tex]m = \frac{115200 - 224000}{16} = -6800[/tex]
Then, the linear model is:
[tex]V(t) = 224000 - 6800t[/tex]
Item b:
The value after 10 years is V(10), so:
[tex]V(10) = 224000 - 6800(10) = 156000[/tex]
Then, the depreciated value of the boat after 10 years is of $156,000.
Item c:
This is t for which: V(t) = 100000. So
[tex]V(t) = 224000 - 6800t[/tex]
[tex]100000 = 224000 - 6800t[/tex]
[tex]6800t = 224000 - 100000[/tex]
[tex]6800t = 124000[/tex]
[tex]t = \frac{124000}{6800}[/tex]
[tex]t = 18.24[/tex]
The depreciated value falls below $100,000 after 18.24 years.
Item d:
The sketch is appended at the end of this answer.
A similar problem is given at https://brainly.com/question/24978772
