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timama [110]
3 years ago
7

The manufacturer's suggested retail price (MSRP) for a particular car is $25,425, and it is expected to be

Mathematics
1 answer:
arsen [322]3 years ago
3 0

Answer:

(a) The linear depreciation function of the car which gives the worth of the car y after a number of years, t is given as follows;

y = 25,425 - 2,739 × t

(b) The value of the car 7 years from now is $6,252

(c) $2,739 per year

Step-by-step explanation:

The manufacture's suggested retail price (MSRP) for the car = $25,425

The amount the car is expected to be worth in 5 years = $11,730

(a) The linear depreciation is given as follows;

Depreciation \ Per \ Year \ = \dfrac{Cost \ of \, Asset -  Salvage \ Value}{Life \ of \, Asset \ in \ use}

Where;

Cost of Asset = $25,425

Salvage Value = $11,730

Life of Asset in use = 5 years

We get;

Depreciation \ Per \ Year \ = \dfrac{\$ 25,425 -  \$11,730}{5} = \$2,739/year

Therefore, the linear depreciation function, can be written as follows;

y - 25,425 = -2,739×(t - 0)

y = -2,739·t + 0 + 25,425 = 25,425 -2,739·t

y = 25,425 - 2,739 × t

Where;

y = The expected worth of the car after a given number of years

t = The number of years used for the calculation of the depreciation

(b) The value of the car 7 years from now is given by substitution as follows;

Whet t = 7, we have;

y = 25,425 -2,739·t = y = 25,425 -2,739 × 7 = $6,252

The value of the car 7 years from now = $6,252

(c) Depreciation \ Per \ Year \ = \dfrac{\$ 25,425 -  \$11,730}{5} = \$2,739/year

The car is depreciating at a rate of $2,739 per year.

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(a) Null Hypothesis, H_0 : \mu \leq 1.4 pounds

    Alternate Hypothesis, H_A : \mu > 1.4 pounds

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Step-by-step explanation:

We are given that the average number of pounds of sliced ham ordered per customer at the deli where Dennis used to work was 1.4, but at his new job, the average of the 32 people who have ordered ham so far is 1.5 pounds, with a standard deviation of 0.3 pounds.

<em>Let </em>\mu<em> = average amount of ham bought at the new store.</em>

So, Null Hypothesis, H_0 : \mu \leq 1.4 pounds     {means that the average amount of ham bought is lesser than or equal to 1.4 pounds at the new store}

Alternate Hypothesis, H_A : \mu > 1.4 pounds     {means that the average amount of ham bought is higher at the new store}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

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So, <u><em>test statistics</em></u>  =  \frac{1.5-1.4}{\frac{0.3}{\sqrt{32} } }  ~ t_3_1  

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The value of t test statistics is 1.886.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of 1.6955 at 31 degree of freedom for right-tailed test.

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Therefore, we conclude that the average amount of ham bought is higher at the new store.

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