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andrew-mc [135]
2 years ago
14

The value of a car decreases $500 for every 1,000 miles it is driven. The current value of the car is $23,000, and the car is dr

iven an average of 10,000 miles per year. What will be the value of the car, in dollars, at a point in time y years from now?
A) $23,000 − $500y

B) $23,000 − $0.02y

C) $23,000 − $5,000y

D) $23,000 − $0.0002y
Mathematics
1 answer:
Hoochie [10]2 years ago
4 0

Answer: C

Step-by-step explanation:

If we know the value of the car decreases $500 for every 1,000 miles, and that the car is driven about 10,000 miles every year, that means that you need to take the total value of the car (23,000) and subtract it from the amount of money it is losing per year. Again, the car is driven about 10,000 miles per year, so that means that the car will most likely continue to be driven 10,000 miles per year. If you do the math, for one year, the value of the car will drop $5,000 ($500 x 10, because it is $500 per every 1,000 miles) So, for each year, you can just multiply the number of years by $5,000 to find out how much the vehicle has depreciated over time.

Hope this helped you and made sense! Feel free to ask me any questions you have!

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pickupchik [31]

This sequence be represented as a recursive equation by a1=8 and an=2a1

<u>Step-by-step explanation</u>:

  • 'Recursive' refers to the repetition of a specific process in a sequence.
  • The given sequence is {8,16,32,64}.
  • If the value is 2 times the previous value, then an=2a(n-1)

Let a1=8,

then a2 = 2a(2-1)

⇒ a2 = 2a1

⇒ a2 = 2(8)

⇒ a2 = 16

Similarly,

For a2=16,

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⇒ a3 = 2(16)

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Congress regulates corporate fuel economy and sets an annual gas mileage for cars. A company with a large fleet of cars hopes to
attashe74 [19]

Answer:

a) Null hypothesis:\mu \leq 30.2  

Alternative hypothesis:\mu > 30.2  

b) X \sim N(\mu=32.12, \sigma=4.83)

And the distribution for the random sample is given by:

\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)

c) t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81    

p_v =P(t_{(49)}>2.81)=0.0035  

d) If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.  

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

\bar X=32.12 represent the sample mean  

s=4.83 represent the sample standard deviation

n=50 sample size  

\mu_o =30.2 represent the value that we want to test

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:  

Null hypothesis:\mu \leq 30.2  

Alternative hypothesis:\mu > 30.2  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

X \sim N(\mu=32.12, \sigma=4.83)

And the distribution for the random sample is given by:

\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)

c. Perform the test and calculate P-value

Calculate the statistic

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

t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=50-1=49  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(49)}>2.81)=0.0035  

d. State your conclusion.

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.  

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

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