Provision * Existing liability * Uncertain timing or amount on future expenditure. (ito ang nagdidistuinguish) * Liability still exist but the date due and amount is indefinite * May equal to estimated liability or loss contingency (accrued because it is both probable and measurable) Provision and other liabilities * P; uncertain OL; certainty of timing and amount Recognition of provision * P shall be recognized in the FS under the ff. conditions
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1. | Question : | The Excel function RAND() generates a random number that is | | | Student Answer: | | between -1 and 0 | | | | between 0 and 1 | | | | between -1 and 1 | | | | between 0 and 100 | | | | Points Received: | 4 of 4 | | Comments: | | | | 2. | Question : | Which of the following Excel formula always generates a number that is between 100 and 200? | | | Student Answer: | | =100 * RAND() | | | | =100 * RAND() + 200
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Binomial Distribution This is a discrete random variable, where the process of obtaining the Binomial distribution is called “Bernoulli “ process. An experiment that often consists of repeated trials, each with two possible outcomes, which could be labeled as “success” or “failure”. This experiment is known as binomial experiment. A binomial experiment is one that possesses the following properties: 1. The experiment consists of n repeated trials. 2. Each trial has only 2 possible
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manufactured company has a mean life time of 700 hours with a standard deviation of 60 hours. Find probability that in random sample of 16 tubes taken out from group have a mean life of between 690 and 710 hours? Has more than 720 hours? Question 3: The masses of 1400 balls are normally distributed with a mean of 21.40 grams and standard deviation of 0.048 grams of 250 random samples of size 34 are drawn from population. Determine the expected mean and standard error of sampling distribution of
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|0.27 |1.00 | |Total |1.00 | | Average material cost per unit = [pic] Simulated values of material cost per unit |Serial Numbers|Random Numbers|Material Cost | |1 |0.6287 |38 | |2 |0.2275 |35 | |3 |0.2757 |35 | |4 |0.7330 |39 |
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IE 361 HOMEWORK 1 In this homework, we have dealt with an inventory problem of an electronic store particularly their mobile phone stocks. We have worked on two alternative scenarios which selling with backordering strategy and without it. Dt=Demand for mobile phones on week t.( Poisson Distribution) Dt={0,1….m} Xt=Inventory level at the end of week t. S=Maximum amount of inventory level. s=Minimum inventory level that does not required new order.(if inventory level strictly
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between breakdowns in Q2. This shows that if the random number falls: between 0 and 0.2 it will take 1 day to repair between 0.2 and 0.65 it will take 2 days to repair between 0.65 and 0.90 it will take 3 days to repair and between 0.90 and 1 it will take 4 days to repair. Using the continuous distribution chart shown, I first set up a table with 52 options for random numbers. I then used the formula =RAND() in Column B to simulate 52 random numbers to then be used for the continuous distribution
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1. Assuming Mr. Jaeger chooses to harvest the grapes before the storm arrives, how much money will he make? If Mr. Jaeger chooses to harvest immediately, he will eliminate the risk, but the quality of wine will be lower than the normal. It results that the price of a bottle of wine will be $2.85, but the quantity of bottled produced will not be changed. The quantity will be 12,000 (1,000 *12=12,000) bottles of wine. Therefore, the revenue he will make will be $34,200 ($2.85*12,000=$34,200)
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confident are you that this answer is a good one? What are the limits of the study? Write at least one paragraph. Answers 1. # of days P(x) Cumulative 1 0.2 0 2 0.45 0.2 3 0.25 0.65 4 0.1 0.9 Q: 2-4. Break Random times b/w Random Repair Random Lost cumulative down # 1 Break (weeks) # 2 Time #3 Revenue time 1 0.78468 5.314929 0.88991 3 2237 $6,711 5.314929 2 0.512227 4.294201 0.831365 2 3244 $6,488 9.60913 3 0.389251 3.743399 0.912647 2 5874 $11,748 13.35253 4 0.998082
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continuous probability distribution function (CDF), we can apply the calculation f(x) = x/18, 0<= x <=6 weeks to determine the weeks between equipment failures. The translation of that function for use in Excel was x=6*SQRT(r), where r is a random number between 0 and 1. I generated pseudorandom numbers for use
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