The largest decrease in size was from 3 to 4, and it leveled out after the 5 to 6 decrease. 2. In Part II of this assignment, you graphed the atomic radius of some elements from group 14 on the periodic table. What is the general trend for atomic radius going down that group? As the period number gets bigger, so does the atomic radius.
1. Factor a trinomial whose leading coefficient is 1. Pick any one of the problems and solve the trinomial. If the trinomial is prime, state this and explain why. a. x2+8x+15 b. x2–4x –5 c. x2–14x+45 2.
TASK 1 A. Complete the attached “Simulation Template” to determine the following costs: 1. Average materials cost per unit The first thing I had to do was figure out the random numbers interval in order to figure out the material cost per unit. Materials Probability Cost Random Numbers Interval* 0.2 $ 33.00 0.00 < 0 .20 0.33 $ 35.00 0.21 < 0.53 0.37 $ 38.00 0.54< 0.90 0.1 $ 39.00 0.91 < 1 *The random number interval came from the probability which was provided in the simulation template. I first start with 0.00 less than the probability which is 0.20 then we start at 0.21 less than 0.53.
A) The bowed shape reflects constant opportunity cost. B) The bowed shape reflects decreasing opportunity cost. C) The bowed shape indicates that opportunity cost at first decreases at a decreasing rate, and then begins to decrease at an increasing rate. D) The bowed shape indicates that opportunity cost at first increases at a decreasing rate, and then begins to increase at an increasing
(TCO 12) Everything else remaining constant, when the sample size _____, sampling error decreases. increases decreases stays constant None of the above 24. (TCO 12) Determine the upper control limit (UCL) and lower control limit (LCL) for an X-Bar chart if six samples of subgroup size 3 (n=3) were collected, the mean of the sample averages is 4.7, and the mean of the sample ranges is 0.35. Determine the upper control limit (UCL) and lower control limit (LCL) for an X-Bar chart. CHART UCL = 4.86905, LCL = 4.53095 UCL = 4.90195, LCL = 4.49805 UCL = 4.72250, LCL = 4.14350 UCL = 5.05805, LCL =
67.029 g to three significant figures b. 0.15 L to one significant figures c. 52.8005 mg to fie significant figures d. 3.174 mol to three significant figures 67.0 g 0.2 L 52.800 mg 3.17 mol 30. State the rules governing the number of significant figures that result from each of the following operations. a. addition and subtraction The answer must have the same number of digits to the right of the decimal point as there are in the measurement that has the fewest digits to the right of the decimal point. b. multiplication and division The answer can have no more significant figures than are in the measurement that has the fewest number of significant figures 31.
K12_1821161 $500 Answer from 3 Jan 9, 2014 1:03:23 PM Page 34. Blackboard Collaborate ?? K12_1821161 $100 Question from 4 Jan 9, 2014 1:03:23 PM Page 35. Blackboard Collaborate ?? K12_1821161 $100 Answer from 4 Jan 9, 2014 1:03:23 PM Page 36.
Year 10 Unit 1 Mathematics Test Total 46 Marks 1. Simplify the following fractions a) 56/84 b) 0.6/1.2 c) 15/45 [3 marks] 2. Change each of these to a mixed number a) 8/3 b) 13/4 c) 17/5 d) 19/7 [4 marks] 3. Calculate these a) 4-2 x 3 b) (4-2) x 3 c) 20 ÷ 1 + 3 [3 marks] 4. Find a) 1.5 % of 50 b) 4.2 % of 500 kg c) 5.7 % of $3000 [3 marks] 5.
Hypothesis Tests Case Problem 1: Quality Associates, Inc. 1. H0: µ = 12 H: µ ≠12 The hypothesis testing results are shown below: Sample 1 Sample 2 Sample 3 Sample 4 Sample Size 30 30 30 30 Mean 11.959 12.029 11.889 12.081 Standard Deviation 0.220 0.220 0.207 0.206 Level of Significance (alpha) 0.010 0.010 0.010 0.010 Critical Value (lower tail) -2.576 -2.576 -2.576 -2.576 Critical Value (upper tail) 2.576 2.576 2.576 2.576 Hypothesized value 12 12 12 12 Standard Error 0.040 0.040 0.038 0.038 Test Statistic -1.027 0.713 -2.935 2.161 p-value 0.2 -0.4 0.4-1 0-0.01 0.01-0.02 Sample 1 calculations (with same calculations for sample 2, 3, and 4): Mean: 368.76/30 = 11.959 Standard Deviation: (11.959 - 11.55)2 + (11.959 - 11.62)2 … (11.959 - 12.15)2 = 1.408 √(1.408/30 - 1) = 0,220 Critical Value: Because = 0.01, /2 = 0.005 = -2.576 (lower tail) and 2.576 (upper tail) Standard Error: 0.22/√30 = 0.04 Test Statistic: 11.959-12/(0.22/√30) = -1.027 p-value: DF:29 and 1.027 is between .2 and.1. These should be timed 2 because its two tailed. Therefore, P is between 0.2 and 0.4. We will reject only sample 3 of the hypothesis H0: µ = 12 because it is operating below the desired mean with a mean of 11.889 and the test statistic of sample 3 is -2.935, which is outside the critical value interval of -.2576 to 2.576.
Why? 3. Suppose the mean cannot be adjusted, but the standard deviation can be reduced. What maximum value of the standard deviation would make 90% of the parts acceptable to the consumer? (Assume the mean continues to be 1.008 inches).