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Keith_Richards [23]
4 years ago
15

A tractor costs $15,450 and depreciates in value by 14% per year. How much will the tractor be worth after 3 years ?

Mathematics
1 answer:
Maru [420]4 years ago
5 0
A=15,450×(1−0.14)^(3)
A=9,827.07
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1,547,489 which digit is in the thousandsths place
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Answer:

7

Step-by-step explanation:

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If f(x)=x+7 and g(x)=1/x-13, What is the domain of (f•g)(x)
astra-53 [7]

Answer:

F(x) = x+7

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The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the
lora16 [44]

Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average sodium content.</em></u>

So, Null Hypothesis, H_0 : \mu = 1100 milligrams      {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, H_A : \mu \neq 1100 milligrams      {means that the sodium content is different from what the nutrition label states}

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

                    T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}  ~ t_n_-_1

where, \bar X = sample average sodium content = 1088.64 milligrams

            s = sample standard deviation = 234.12 milligrams

            n = sample of packages of Oriental Spice Sauce = 40

So, <u><em>test statistics</em></u>  =  \frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}  ~ t_3_9

                              =  -0.307

The value of z test statistics is -0.307.

<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.</em><em> </em>

<em>Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the sodium content is same as what the nutrition label states.

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