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N76 [4]
3 years ago
15

in the year 2001, a person bought a new car for $22000. for each consecutive year after that, the value of the car depreciated b

y 5%. how much would the car be worth in the year 2005, to the nearest hundred dollars
Mathematics
1 answer:
igomit [66]3 years ago
7 0

Answer:

To the nearest hundred dollars, the car will be worth $17,900 by 2005

Step-by-step explanation:

Firstly, we need to write the depreciation equation

We have this as:

V = I(1 - r)^t

V is the present value which is what we want to calculate

I is the initial value, the amount the cad was bought which is $22,000

r is the rate of change which is 5% = 5/100 = 0.05

t is the time difference which is 2005-2001 = 4

Substituting all these into the depreciation equation, we have it that

V = 22,000(1 -0.05)^4

V = $17,919.1375

To the nearest hundred dollars, that would be;

$17,900

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LUCKY_DIMON [66]
Let success be that a plant dies in the winter and failure be that a plant survives the winter, then
 p = 0.05, q = 1 - 0.05 = 0.95

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In which quadrant does the given point lies??? (3,-2)
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Quadrant 4 .
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4 years ago
Which linear inequality is represented by the graph?
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The correct answer is B. I figured this out by testing out coordinates into each equation. I am only going to show my work for equation B though... if you want to see the rest of my work, message me.

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AREAS AND VOLUMES OF SIMILAR SOLIDS URGENT?
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Two similar prisms have proportional dimensions. Find the ratio between lengths of these prisms:

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Answer: 137.2 cub. m.

6 0
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