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lulu5351
11.01.2020 •
Mathematics
Having some trouble , can anyone me out ?
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Ответ:
a)![y=-3950\cdot{x}+21500](/tpl/images/0455/8064/6ca82.png)
b) f(x)![=21500\cdot{(0.7953)^x}](/tpl/images/0455/8064/a7399.png)
c) Year 1: Linear $17550, exponential $17099
Year 4: Linear $5700, exponential $8601
d) Exponential model
e) The linear model depreciates the value quicker than exponential model long term around 4 years
Step-by-step explanation:
a) At year 0 the price is 21500 and at year 2 the price is 13600
WE can use points (0,21500) and (2,13600)
We can determine the gradient
We can use the point slow formula:
b) We can use the following equation:
f(x)![=a\cdot{(1+r)^x}](/tpl/images/0455/8064/c7194.png)
f(x) is the depreciation value after amount of time t, a is the new value, r is the rate of depreciation and x is the time.
The depreciation rate is 20.47% and is negative because it decreases the new value of the car
c) Year 1:
f(x)![=21500\cdot{(0.7953)^1}=17099](/tpl/images/0455/8064/56379.png)
Year 4:
f(x)![=21500\cdot{(0.7953)^4}=8601](/tpl/images/0455/8064/df0e3.png)
d) Year 2
f(x)![=21500\cdot{(0.7953)^2}=13598](/tpl/images/0455/8064/bd513.png)