QUESTION IMAGE
Question
select the correct answer from each drop-down menu.
tonya and leo each bought a cell phone at the same time.
the trade-in values, in dollars, of the cell phones are modeled by the given functions, where x is the number of months that each person has owned the phone
tonya’s phone: ( f(x) = 490(0.90)^x )
leo’s phone:
| x | g(x) |
|---|---|
| 2 | 360 |
| 4 | 270 |
____ phone has the greater initial trade-in value
during the first four months, the trade-in value of tonya’s phone decreases at an average rate ____ the trade-in value of leo’s phone
reset next
To solve the problem, we analyze the initial trade - in values and the average rate of decrease for both phones.
Step 1: Find the initial trade - in values
- For Tonya's phone, the function is \(f(x)=490(0.90)^{x}\). The initial trade - in value occurs when \(x = 0\). Using the property of exponents \(a^{0}=1\) (where \(a = 0.90\) in this case), we have \(f(0)=490\times(0.90)^{0}=490\times1 = 490\) dollars.
- For Leo's phone, we look at the table. When \(x = 0\), \(g(0)=400\) dollars.
Since \(490>400\), Tonya's phone has the greater initial trade - in value.
Step 2: Calculate the average rate of decrease for Tonya's phone
The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is given by the formula \(\frac{f(b)-f(a)}{b - a}\).
For Tonya's phone, we want to find the average rate of decrease over the first 4 months, so \(a = 0\) and \(b = 4\).
First, find \(f(4)\):
\(f(4)=490\times(0.90)^{4}\)
\((0.90)^{4}=0.90\times0.90\times0.90\times0.90 = 0.6561\)
\(f(4)=490\times0.6561=321.489\)
Now, use the average rate of change formula:
\(\frac{f(4)-f(0)}{4 - 0}=\frac{321.489 - 490}{4}=\frac{- 168.511}{4}\approx - 42.13\)
Step 3: Calculate the average rate of decrease for Leo's phone
For Leo's phone, using the table, when \(x = 0\), \(g(0)=400\) and when \(x = 4\), \(g(4)=270\).
The average rate of change over the interval \([0,4]\) is \(\frac{g(4)-g(0)}{4 - 0}=\frac{270 - 400}{4}=\frac{-130}{4}=- 32.5\)
Since \(\vert - 42.13\vert>\vert - 32.5\vert\), the trade - in value of Tonya’s phone decreases at an average rate greater than the trade - in value of Leo’s phone.
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s:
- The phone with the greater initial trade - in value: Tonya's
- The relationship between the average rate of decrease of Tonya's phone and Leo's phone: greater than