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Sergio039 [100]
3 years ago
14

In efforts of improving their orofit margins, a company is considering creating their own manufacturing plant for their product.

It had a start-up cost of 25,000 to build the plant. This plant will drop the production cost down to $15 per item. If each item sells for $50, then how many items would it take to get an overall profit of $100,000 for the year?
Mathematics
1 answer:
Trava [24]3 years ago
7 0

Answer:

To get an overall profit of $100,000, then 3,571 items is needed to be produced

Step-by-step explanation:

Mathematically;

Profit = Revenue - Production cost

In the question, we are told that the profit we are expecting is $100,000

Let the number of items to be produced for sales be x items

Now let’s look at the production cost ;

The production cost will be number of items for sale * production cost = 15 * x = $15x

Now, we need to add the start-up cost to the production cost to get the total amount spent on production;

Mathematically the total amount would be = $25,000 + $15x

Let’s look at the revenue;

The revenue will be selling price of each item * price of each item;

Mathematically, that would be 50 * x = $50x

So therefore;

50x - (25,000 + 15x) = 100,000

50x -25,000 -15x = 100,000

35x = 100,000 + 25,000

35x = 125,000

x = 125,000/35 = 3571.42 which is approximately 3,571 items

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Answer:

  • There is no significant evidence that p1 is different than p2 at 0.01 significance level.
  • 99% confidence interval for p1-p2 is  -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

H_{0}: p1-p2=0

H_{a}: p1-p2≠0

Test statistic can be found using the equation:

z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

  • p1 is the sample proportion of the common attribute in population1 (\frac{16}{30} =0.533)
  • p2 is the sample proportion of the common attribute in population2 (\frac{1337}{1900} =0.704)
  • p is the pool proportion of p1 and p2 (\frac{16+1337}{30+1900}=0.701)
  • n1 is the sample size of the people from population1 (30)
  • n2 is the sample size of the people from population2 (1900)

Then z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}} ≈ 2.03

p-value of the test statistic is  0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2±z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}} where

  • z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704±2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}} ≈ -0.171 ±0.237 that is (−0.408, 0.066)

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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