Why or why not? | Not unusual, because it is within 2 standard deviations of the mean | 8d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98°F or lower? | 0.62 percent | 9a.
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: ____Convert 0.2 to a % by times by 100. So 0.2 = 20%, so 20% of the data is between 50 and 59______________ 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.
6. Did the empowerment variable or self-care self-efficacy variable demonstrates the greatest amount of dispersion? Provide a rationale for your answer. Ans: The experimental group's mean self-care self-efficacy posttest score (mean = 96.00) was 6.44 points higher than its baseline mean score (mean = 89.56). In total, there is a
I first start with 0.00 less than the probability which is 0.20 then we start at 0.21 less than 0.53. I came up with 0.53 from adding the initial probability of 0.20 and the second 0.33. Next I start at 0.54 less than 0.90 these figures come from adding 0.53 plus the third probability of 0.37. Lastly, I start at 0.91 less than 1 and again I adding 0.90 plus the last probability of 0.10. Now let’s find the Average materials cost per unit.
Chapter 6 8. 1. The parameters of the opportunity set are: E(rS) = 15%, E(rB) = 9%, sS = 32%, sB = 23%, r = 0.15, rf = 5.5% From the standard deviations and the correlation coefficient we generate the covariance matrix [note that Cov(rS, rB) = rsSsB]: Bonds Stocks Bonds 529.0 110.4 Stocks 110.4 1024.0 The minimum-variance portfolio proportions are: wMin(B) = 0.6858 The mean and standard deviation of the minimum variance portfolio are: E(rMin) = (0.3142 × 15%) + (0.6858 × 9%) = 10.89% = [(0.31422 × 1024) + (0.68582 × 529) + (2 × 0.3142 × 0.6858 × 110.4)]1/2 = 19.94% % in stocks % in bonds Exp. return Std dev. 00.00 100.00 9.00 23.00 20.00 80.00 10.20 20.37 31.42 68.58 10.89 19.94 Minimum variance
1a) Calculate molar solubility of SrF2 in: 0.010 M Sr(NO3)2 SrF2 <=> Sr+2 & 2 F- Ksp = [Sr+2] [F-]^2 4.3x10^-9 = [0.010 M] [F-]^2 [F-]^2 = (4.3x10^-9) / [0.010 M] [F-]^2 = 4.3x10^-7 [F-] = 6.56 X 10^-4 Molar by the equation: 1 mole of SrF2 <=> Sr+2 & 2 moles of F- the molar solubility of SrF2 that releases [F-] is half as much: 3.279 X 10^-4 Molar SrF2 yoiur answer, rounded to 2 sig figs would be 3.3 X 10^-4 Molar SrF2 ======================================… 1b) Calculate molar solubility of SrF2(Ksp=4.3x10^-9) in 0.010 M NaF Ksp = [Sr+2] [F-]^2 4.3x10^-9 = [Sr+2] [0.010]^2 [Sr+2] = 4.3x10^-9 / [0.010]^2 [Sr+2] = 4.3x10^-9 / 1 X 10^-4 [Sr+2] = 4.3x10^-5 Molar by the equation:
1 5.24 mL 10.061 g 2 5.49 mL 9.293 g 3 5.99 mL 8.055 g 4 6.51 mL 6.647 g 5 7.02 mL 5.189 g 6 7.51 mL 3.932 g Calculation: Error: Our percent error was -0.37%. % error = EV-AV * 100 % error = 2.69-2.70 * 100 = -0.37% AV
Concerning the representation of racial and ethnic minorities among American psychologists, which of the following statements is TRUE? a. The number of minority members entering the field is about the same as it was 10 years ago. b. The number of minority members entering the field is higher than it was 10 years ago.
Was there a normal distribution? Defend your position by interpreting the normality tests. (0 marks as this question is part of data screening for the writing of the results in Task 10) The results show that the test is not significant as according to Smirnov and Shapiro-Wilk, the sig. =.200, .175, .183 and .122, which are greater than .05. Thus meaning that the assumption of normality was met According to skewness and kurtosis there is no problem with normality as both scores of -1.57 (skewness) and -0.39 (kurtosis) are both within the range of +/- 3.29.
Are they the same? Would you expect them to be about the same? Why or why not? The traffic sent and received for the PK512_100Mbps are pretty much right on. I would believe them to be the same.