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erma4kov [3.2K]
3 years ago
15

In 2012 your car was worth $10,000. In 2014 your car was worth $8,850. Suppose the value of your car decreased at a constant rat

e of change. Define a function f to determine the value of your car (in dollars) in terms of the number of years t since 2012.
Mathematics
1 answer:
monitta3 years ago
7 0

Answer:

The function to determine the value of your car (in dollars) in terms of the number of years t since 2012 is:

f(t) = 10000(0.9407)^t

Step-by-step explanation:

Value of the car:

Constant rate of change, so the value of the car in t years after 2012 is given by:

f(t) = f(0)(1-r)^t

In which f(0) is the initial value and r is the decay rate, as a decimal.

In 2012 your car was worth $10,000.

This means that f(0) = 10000, thus:

f(t) = 10000(1-r)^t

2014 your car was worth $8,850.

2014 - 2012 = 2, so:

f(2) = 8850

We use this to find 1 - r.

f(t) = 10000(1-r)^t

8850 = 10000(1-r)^2

(1-r)^2 = \frac{8850}{10000}

(1-r)^2 = 0.885

\sqrt{(1-r)^2} = \sqrt{0.885}

1 - r = 0.9407

Thus

f(t) = 10000(1-r)^t

f(t) = 10000(0.9407)^t

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Data given

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\bar X_{2}=770 represent the mean for sample 2  

s_{1}=67 represent the sample standard deviation for 1  

s_{2}=56 represent the sample standard deviation for 2  

n_{1}=13 sample size for the group 2  

n_{2}=19 sample size for the group 2  

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Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

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We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=13+19-2=30  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

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Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

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