State the place value of the underlined digit in 6 953 742. A Hundreds C Ten thousands B Thousands D Hundred thousands ( 3. Round off 5 987 341 to the nearest hundred thousand. A 5.8 million C 6.0 million ( B 5.9 million D 6.1 million 4. Which of the following numbers, when rounded off to the nearest thousand, becomes 7 541 000?
None b. i only c. ii only d. i and ii e. ii and iii 5. For the positive integers a,b, and k, ak || b means that ak is a divisor of b, but ak+1 is not a divisor of b. If k is a positive integer and 2k || 72, then k is equal to p. 2 q. 3 r. 4 s. 8 t. 18 6. If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p?
Significant Figures/Scientific Notation Question 1 How many significant figures are in the following values? a. 4.02 x 10-9 b. 0.008320 c. 6 x 105 d. 100.0 Question 2 How many significant figures are in the following values? a.
The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2. What does the 0.2 value indicate? Answer: 0.2 is equal to 1/5 or 20%, 0.2 indicates 20% of the data values are in this particular interval. 3. When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6.
A divide by 100 B divide by 1000 C multiply by 100 D multiply by 1000 15 Which of the following is the smallest? A 0.07 7 C ----10 1 B -7 D 73% 7 A ----12 7 C ----16 7 B -9 7 D -8 5 Which one of these decimals could be rounded to 34.71? A 34.715 B 34.7 C 34.707 D 34.7039 1 6 What is -- as a percentage? 6 A 16% B 1.6% -C 16 2 % 3 D 60% 7 What is 37.5% of one day? A 10 hours B 8.6 hours C 9.6 hours D 9 hours 8 Increase $38 by 15%.
C. Sum of the possible outcomes is 1.00. D. The outcomes are mutually exclusive. Ans: A 6) In which of the following distributions is the probability of a success usually small? A. Binomial B. Hypergeometric C. Poisson D. All distribution Ans: C 7) The difference between the sample mean and the population mean is called the A. Population mean.
507 B. 607 C. 617 4. Which sentence is true? A. 48 – 23 is less than 43 – 28.
Since 2.9134>1.2816, 1.6449 and 2.3263, I rejected H0 and accepted Ha at 1 = 0.10, 0.05, and 0.01 and concluded that the mean rating exceeds 42. Since 2.9134<3.0902, I rejected H0 at a = 0.001, and fail to conclude that the mean rating exceeds 42. c. Using the information in part b, calculate the p-value and use it to test H0 versus Ha at each of at = 10, .05, .01 and .001. Upper tail p-value for z = 2.9134 is 00018. Since 0.0018<0.10, 0.65, 0.01, I rejected H0 and accepted Ha at a = 0.10, 0.05, and 0.01 and concluded that the mean rating exceeds 42. Since 0.0018>0.001, I failed to reject H0 @ a = 0.001 and failed to conclude that the mean rating exceeds 42. d. How much evidence is there that the mean composite
0.36 B. 2.3 C. 9.0 D. 11 (1) 3. What is ∆H for the reaction below in kJ? CS2(g) + 3O2(g) ® CO2(g) + 2SO2(g) [∆Hf / kJ mol–1: CS2(g) 110, CO2(g) – 390, SO2(g) – 290] A. −570 B.
Harmon Foods Case 1) Case Shipments = -80,315.68 + 946.56T + 3877.95S + .096C + .076D - .048CL - .02DL-2 Where T - Time in months since Dec 1983 S - Seasonal Index as given by the case C - Consumer Pack spending for the current month (packs*$.2 discount *24 packages/case) D - Dealer Allowance for the current month CL- Consumer Pack spending lagged from 1 month ago (packs*$.2 discount *24 packages/case) DL-2- Dealer Allowance lagged from 2 months ago We assume that one consumer pack includes 24 coupons (a case) at a value of $.20/ea. Based on information in the case, we weighted the consumer packs and dealer allowance to be equal to 90% of the current month and 5% from each of the two previous months. Also, we assumed that these allowances would immediately increase sales and then decrease future sales because consumers and dealers would stock up during canvassing periods. Therefore, we included values for lagged effects of previous consumer packs and dealer allowances. 2) Case Shipments = -80,315.68 + 996.56*49 + 3877.95*113 + .096*960,000 + .076*154,455 - .048*345,028.8 - .02*368,337.5 = 483,432 Standard Error = 34,733.12 given by Excel Confidence Interval of 95% -> T Value of 97.5% for 42 degrees of freedom is 2.018 2.018 * 34,733.12 * SQRT (1 + (1/48)) = 70,818 95% Confidence Interval = Estimate +/- Standard Error*T-Value*SQRT(1+(1/n)) The 95% Confidence Interval is from 412,614 and 554,250 cases.