Mathematics, 10.04.2020 18:46 and7735

Olivia purchased a car worth $18,000 in the year 2001. It loses its value by 6% per year. What is the value of the car in 2004? Round to the nearest cent.

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The value of car in 2004 is $14951, if Olivia purchased a car worth $18,000 in the year 2001 and it loses its value by 6% per year.

Step-by-step explanation:

The given is,

Car worth $18,000 in the year 2001

Loses its value by 6% per year

Step:1

Formula to calculate the worth of car with loss rate,

$y = A(1-r)^{t}$ .....................................(1)

Where, y = Value of car after t period

A = Initial value of car

r = Loss rate

t = Time period

From give,

A = $18,000

r = 6 % per year

t = 3 years

Equation (1) becomes,

$y = 18000(1-0.06)^{3}$

$=18000(0.94)^{3}$

= (18000 × 0.830584)

= 14950.512

≅ 14951

Value of car in 2004, y = $14951

( or )

Step:1

t = 3 years ( 2001 to 2004 )

For 1st year,

= ( Worth of car in previous year ) -

( Worth of car in previous year × Loss rate )

= ( 18000 ) - ( 18000 × 0.06 )

= 18000 - 1080

= $16920

For 2nd year,

= ( 16920 ) - ( 16920 × 0.06 )

= 16920 -1015.2

= $15904.8

For 3rd year,

= ( 15904.8 ) - ( 15904.8 × 0.06 )

= 15904.8 - 954.288

= 14950.512

≅ $14951

Result:

The value of car in 2004 is $14951, if Olivia purchased a car worth $18,000 in the year 2001 and it loses its value by 6% per year.

Answer from: Quest

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step-by-step explanation:

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Answer from: Quest

i dont know

step-by-step explanation:

What intervals is the function increasing? decreasing?

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Olivia purchased a car worth $18,000 in the year 2001. It loses its value by 6% per year. What is th...

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