up costs: 12‚200 Cost driver: number of set ups. Total number of set ups: 5 (12‚200/5) x 2 = 4‚880 (12‚200/5) x 2 = 4‚880 (12‚200/5) x 1 = 2‚440 Purchasing and checking wood edging costs: 8‚000 Cost driver: wood edging per cabinet Total number of meter per wood edging: 8 meters (8‚000/8) x 4 = 4‚000 (8‚000/8) x 4 = 4‚000 (8‚000/8) x 0 = 0 Purchase of wood costs: 5‚100 Cost driver: wood (meters) per cabinet Total number of meter of wood: 40 meters
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Factor Rating (Facility Location) Example-01 A manufacturer of garments is actively considering five alternative locations for setting up its factory. The locations vary in terms of their advantages to the firm. Hence‚ the firm requires a method of identifying the most appropriate location. Based on a survey of its senior executives‚ the firm has arrived at six factors to be considered for final site selection. The rating of each factor on a scale of 1 to 100 provides this information. Further
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Question 4. Graph from Excel From the graph – it can be seen that as the tosses number increase‚ so also the head proportion moving toward 0.5 In our simulation‚ the first toss has an H‚ which we can see because the proportion starts at 1.0. Following the first simulated toss‚ the proportion changes with each new toss depending on whether we got an H or a T. Answer 5 Each time that a new set of random numbers is generated‚ the graph changes accordingly. However‚ in all of these graphs we can
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Some sample solutions of annuity problems Typical MAT 112 problems: What is the value of an ordinary annuity at the end of 10 years if $300 is deposited each quarter into an account earning 7 % compounded quarterly? Also‚ of this total value‚ how much did you contribute and how much is from the interest? For 40 deposits of $300 each with [pic]‚ we find the accumulated value as [pic] The total interest earned is the difference between the amount in the account and what you actually
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D) x3 +8 x 3 + 4x 2 − 8x + 8 x 3 − 4x 2 + 8x + 8 x 3 + 8x + 8 13. The difference of twice a number and six is four times the number. Find an equation to solve for the number. A) 2x – 6 = 4 B) 2x – 6 = 4x C) 2x + 6 = 4x D) 2x – 6 = x + 4 14. Expand: A) 4 x 2 B) 4 x 2 C) 2 x 2 D) 4 x 2 (2 x − 3) 2 −9 +9 − 12 x + 9 − 12 x + 9 CPT Review 4/17/01 3 15. Which of the following numbers is the smallest? 3 A) − 4 3 B) − 2 C) − 1 2 D) − 3 16. Which of the following is the largest? A)
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section. In order to write this program‚ two variables‚ ROWNUM and COLUMNNUM‚ will be used with nested repetition loop to complete the necessary calculations. ROWNUM will be used in an outerloop and contain numbers between1-10 and COLUMNNUM will be used in an innerloop and contain numbers between 1-10. For each iteration of the outerloop‚ the inner loop will be executed 10 times and the product of ROWNUM*COLUMNNUM will produce the multiplication table of ROWNUM. The ROWNUM and COLUMNNUM variables
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JOHOR EDUCATION DEPARTMENT MINISTRY OF EDUCATION MALAYSIA ADDITIONAL MATHEMATICS PROJECT WORK 2013 SOLUTION OUTLINE PART A Monthly Income (RM) Number of Family Members Categories Allocation of Income (RM) Allocation of Income (%) 3000 6 Food 900.00 30 Utility 660.00 22 Transportation 450.00 15 Education 300.00 10 Recreation 150.00 5 Others 540.00 18 Total 3000.00 100 (i) Example : Pie Chart‚ Bar Chart and Line Graph. (ii) Mean METHOD 1 = RM
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9. Owls eat small mammals. They regurgitate the bones and fur in balls called pellets. The table shows the contents of 62 pellets from long-eared owls. Number of mammals found in the pellet Frequency 1 9 2 17 3 24 4 6 5 5 6 1 (a) Show that the total number of mammals found is 170 1 mark (b) Calculate the mean number of mammals found in each pellet. Show your working and give your answer correct to 1 decimal place. 2 marks (c) There are about 10 000 long-eared owls
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MCB 150 Lab PROBLEM SET 1 1. The data below were obtained after performance of Ex. 2 as stated in your MCB 150 lab manual. a. Estimate the number of total bacteria‚ actinomycetes‚ fungi and algae per gram (dry weight) of soil collected from the IBS garden pH: 6.0 ODW: 8.6 grams Colony Counts: PDA 10-1: 210‚ 190 AMA 10-4: 125‚ 119 SCA 10-4: 80‚ 62 10-2: 20‚ 34 10-5: 17‚ 29 10-5: 18‚ 2 Algal Medium (no. of + tubes/ no. of inoculated tubes): 10-1: 5/ 5 10-2: 5/ 5 10-3: 3/ 5 10-4:
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| |A) |Hours spent preparing for an exam is a dependent variable. | |B) |There is no relationship between the number of hours spent preparing for the exam and the average grade.
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