Sofia purchased a new car in 2007 for$34,000. The value of the car has been depreciating exponentially at a constant rate. If the value of the car was $16,200 in the year 2014, then what would be the predicted value of the car in the year 2021, to the nearest dollar?

$7,718

Step-by-step explanation:

Given that :

Cost of car in 2007 = $34000

Cost of car in 2014 = $16200

Predicted value of car in 2021

Lets obtain the percentage Depreciation (d) :

Worth (A) of car in 2014 = $16200

Time (t) = 2014 - 2007 = 7 years

Initial cost (I) = price in 2007

A = I(1 - d/100)^7

16200 = 34000( 1 - d/100)^7

16200 = 34000(1 - 0.01d)^7

(16200/34000)^1/7= 1 - 0.01d

0.8995 = 1 - 0.01d

0.01d = 1 - 0.8995

d = 0.1005 /0.01

d = 10.05

Value of car in year 2021:

t = 2021 - 2007 = 14 years

A = I(1 - 10.05/100)^14

A = 34000(1 - 0.1005)^14

A = 34000(0.8995)^14

A = 34000 * 0.2269950

A =7717.83100

Approximate worth = $7718

3

step-by-step explanation:

because where the point where the 2 lines intersect is (3,-1) the x value is 3

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.