Car cost $40000 it depreciates 12% each year. What is the cars value in 5yrs

1 answer:

8 0

The formula is
A=p (1-r)^t
A future value?
P present value 40000
R rate of depreciation 0.12
T time 5years
A=40,000×(1−0.12)^(5)
A=21,109

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Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 in

nadezda [96]

Answer:

$z=-1.99$

$z=1.99$

And if we solve for a we got

$a=288 -1.99*3.7=214.4$

$a=288 +1.99*3.7=295.4$

And the limits for this case are: (214.4; 295.4)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

$X \sim N(288,3.7)$

Where $\mu=288$ and $\sigma=3.7$

The confidence level is 95.44 and the signficance is $1-0.9544=0.0456$ and the value of $\alpha/2 =0.0228$ . And the critical value for this case is $z = \pm 1.99$