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gayaneshka [121]
2 years ago
13

Suppose that a small family farm sold its output for $100,000 in a given year. The family spent $25,000 on fuel, $40,000 on seed

, fertilizer, and pesticides, and $25,000 on equipment, including maintenance. The family members could have earned $20,000 working at other occupations.
What is the economic profit at the family farm?
Mathematics
2 answers:
zheka24 [161]2 years ago
8 0
First we have to evaluate our income: $100,000. And then the expenditure (sum of all expenses): $25,000 + $40,000 + $25,000 = $90,000. This is the expenditure, taken away from our income we get $10,000, this is the economic profit of the family farm.
Viefleur [7K]2 years ago
8 0

Answer:

The economic profit is $10,000.

Step-by-step explanation:

A small family farm sold its output for $100,000

The family spent on fuel = $25,000

The family spent on seed fertilizer and pesticides =  $40,000

The family spent on equipment and maintenance = $25,000

They spent total = 25,000 + 40,000 + 25,000

                           = $90,000

The economic profit at the family farm = 100,000 - 90,000

                                                                =  $10,000

The economic profit is $10,000.

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The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv
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Answer:

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=14.42 represent the sample mean

\mu population mean (variable of interest)

\sigma=6.95 represent the population standard deviation

n=36 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

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