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Effectus
3 months ago
10

A new car is purchased for 20300 dollars. The value of the car depreciates at 9.5% per year. What will the value of the car be,

to the nearest cent, after 11 years?
Mathematics
1 answer:
Svet_ta [12.7K]3 months ago
6 0
The initial value stands at 20,300, decreasing annually by 9.5%.

Given that it decreases by 9.5% each year based on the preceding amount, we can apply an exponential decay model.

A 9.5% reduction means multiplying by 91.5% every year.

We express this mathematically. Plugging in 11 years for t yields.

A(t)=23000(0.915)^{t}\\\\A(11)=23000(0.915)^{11}\\\\A(11)=7671.18

$7,671.18
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Answer:

a) Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

b) p_v =P(t_{17}

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c) p_v =P(t_{17}

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d) p_v =P(t_{17}

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

\bar X represents the average of the samples

s denotes the standard deviation of the samples

n=18 indicates the number of samples

\mu_o =10 is the value we are examining

\alpha defines the significance level for the test.

t represents the specific statistic of interest

p_v indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

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

Part b

In this scenario t=-2.3, \alpha=0.05

Initially, we need to calculate the degrees of freedom df=n-1=18-1=17

Since this is a left-tailed test, the p-value is determined by:

p_v =P(t_{17}

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8, \alpha=0.01

We need to find the degrees of freedom df=n-1=18-1=17

For the left-tailed test, the p-value is given by:

p_v =P(t_{17}

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6, \alpha=0.05

Firstly, we find the degrees of freedom df=n-1=18-1=17

Since we are conducting a left-tailed test, the p-value is calculated as:

p_v =P(t_{17}

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

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g A sailing ship stopped at several islands, releasing one mating pair of goats onto each one. On the large island, the goats' g
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Response:

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Detailed explanation:

After t years, the population of goats is described by

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t = \dfrac{6.2}{0.5} = 12.4

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