Question

part 1 of 4 (a) how much will a computer be worth 1 yr after purchase? one year after the purchase, the computer will be worth $800. part 2 of 4 (b) after how many years will the computer be worth only $200? after yr, the computer will be worth only $200. part 3 of 4 (c) determine the y - intercept and interpret its meaning in the context of this prob the y - intercept is ( , ) and it represents the select value.

Explanation:

1. Assume a linear - depreciation model:

• Let the value of the computer \(V\) be a linear function of time \(t\) (in years) of the form \(V(t)=mt + b\). We know that when \(t = 1\), \(V(1)=800\). But we need another piece of information to fully determine the function. Let's assume the computer has a scrap - value of \(0\) after a certain number of years. For simplicity, if we assume straight - line depreciation and assume the computer has an initial value \(V(0)=1000\) (we can assume this for the sake of creating a linear model, since the problem is incomplete without some initial - value information. In a real - world scenario, we would need more data). The general formula for straight - line depreciation is \(V(t)=V_0-rt\), where \(V_0\) is the initial value and \(r\) is the rate of depreciation per year.
• If \(V_0 = 1000\) and \(V(1)=800\), then we can find the rate of depreciation \(r\).
• Substitute into the formula \(V(t)=V_0-rt\):
• When \(t = 1\) and \(V(1)=800\), \(V(1)=V_0-r\times1\). So, \(800 = 1000 - r\), which gives \(r = 200\) per year. So the value function is \(V(t)=1000 - 200t\).

1. Answer part (b):

• We want to find \(t\) when \(V(t)=200\).
• Set \(V(t)=200\) in the equation \(V(t)=1000 - 200t\).
• \(200=1000 - 200t\).
• First, add \(200t\) to both sides: \(200 + 200t=1000\).
• Then subtract \(200\) from both sides: \(200t=1000 - 200=800\).
• Divide both sides by \(200\): \(t=\frac{800}{200}=4\).

1. Answer part (c):

• For the linear function \(V(t)=1000 - 200t\), in the form \(y = mx + b\) (where \(y = V(t)\) and \(x = t\)), the \(y\) - intercept occurs when \(t = 0\).
• When \(t = 0\), \(V(0)=1000\). So the \(y\) - intercept is the point \((0,1000)\).
• In the context of this problem, the \(y\) - intercept represents the initial value of the computer at the time of purchase.

Step1: Determine the depreciation function

Assume initial value \(V_0 = 1000\), using \(V(1)=800\) to find rate \(r = 200\), so \(V(t)=1000 - 200t\).

Step2: Solve for \(t\) in part (b)

Set \(V(t)=200\) in \(V(t)=1000 - 200t\), get \(200=1000 - 200t\), then solve for \(t\) as \(t = 4\).

Step3: Find the \(y\) - intercept

For \(V(t)=1000 - 200t\), when \(t = 0\), \(V(0)=1000\), so \(y\) - intercept is \((0,1000)\) and it's the initial value.

Answer:

(b) 4
(c) \((0,1000)\), initial