80 × 2 = 160 160 × 2 = 320 320 × 2 = 640 The next three terms of the sequence are 160, 320, and 640. 6. 100, 50, 25, … SOLUTION: Calculate common ratio. The common ratio is 0.5. Multiply each term by the common ratio to find the next three terms.
Now we all the 12 costs per units amount from table 2 and divided by 12 in order to get the average. 38 + 38 + 38 + 35 + 35 + 38 + 35 + 38 + 33 + 39 +35 +35 = 437 Now 437/12 = 36.45 The average materials cost per unit is $36.45 2. Average labor cost per unit I use the same formula as I did above to figure out the average labor cost per unit. Labor Probability Cost Random Numbers Interval 0.18 $ 22.00 0.0 < 0.18 0.08 $ 23.00 0.19 < 0.26 0.25 $ 24.00 0.27 < 0.51 0.19 $ 25.00 0.52 < 0.70 0.3 $ 28.00 0.71 <
On page 29 is the explanation of uncertainty. The concept is illustrated in Figure 13 on page 30. I expect your report will include length to 0.1 cm, a whole mm, and to 0.001 m. As an example, a sheet of paper is 21.6 cm, 216 mm, or 0.216 m wide. In each case, the measurement is precise to 1 mm (which is 0.1 cm and 0.001 m). The first degree of uncertainty is at the mm length.
Practice Problems Psy 315 Chapter 2: 11. The mean is 2, the median is 2, the sum of squared deviations is -52, the variance is -2.47, and the standard deviation is 1.57. 12. The mean is 1.361, the median is 1.3124, the sum of squared deviations is 0.07608920.0190223, the variance is 0.0190223, and the standard deviation is 0.1379213544. 13.
We can conclude that the data are Poisson distributed. Chi-Square test of independence Problem 12.12 Use the following contingency table to determine whether variable 1 is independent of variable 2. Let α = .01 | Variable 2 | Variable1 | 24 | 13 | 47 | 58 | | 93 | 59 | 187 | 244 | Step 1 Ho: the two classifications are independent Ha: the two classifications are dependent Step 2 d.f = (r – 1) (c – 1) Step 3 α = 0.01 x 2 0.01, 3df = 11.3449 Step 4 Reject Ho if x 2 > 11.3449 | Variable 2 | Total | Variable1 | 24 (22.92) | 13 (14.10) | 47 (45.83) | 58 (59.15) | 142 | | 93 (94.08) | 59 (57.90) | 187 (188.17) | 244 (242.85) | 583
So the lease term was 3 years, and the useful life of the equipment is 4 years. This indicated that the lessee has use the major portion of the expected economic life of the asset. b) The present value of the minimum lease payments is equal to or greater than “Substantially all” of the fair value of the asset. This rule means that since the fair market value (FMV) at the lease inception is $265,000 and the lessee has used the equipment as its lease obligation of $244,370 this
1,054,848 c. 1,405,888 d. 1,045,828 5. Evaluate: 12xy, when x = 8 and y = 11. a. 31 b. 228 c. 1056 d. 188 6. Write
1. The graph approximates the points: E(r) σ Minimum Variance Portfolio 10.89% 19.94% Tangency Portfolio 12.88% 23.34% 10. The reward-to-variability ratio of the optimal CAL (using the tangency portfolio) is: 11. a. The equation for the CAL using the tangency portfolio is: Setting E(rC) equal to 12% yields a standard deviation of: 20.56% b. The mean of the complete portfolio as a function of the proportion invested in the risky portfolio (y) is: E(rC) = (l - y)rf + yE(rP) = rf + y[E(rP) - rf] = 5.5 + y(12.88 - 5.5) Setting E(rC) = 12% ==> y = 0.8808 (88.08% in the risky portfolio) 1 - y = 0.1192 (11.92% in T-bills) From the composition of the optimal risky portfolio: Proportion of stocks in complete portfolio = 0.8808 × 0.6466 = 0.5695 Proportion of bonds in complete portfolio = 0.8808 × 0.3534 = 0.3113 12.
Well, first of all Stanley did not have to do taxes because his income is way to low and he could have chosen not to file taxes. But still, Stanley’s taxable income would be as follows. His AGI is$3000, which are $1000 +$2000. He did not have any itemized deductions, so we will assume there is standard deduction, which is 5700 for single filers. He is claimed as a dependant on other tax return, so we cannot deduct him as an exemption.
Descriptive Statistics: Income ($1000), Size, Credit Balance ($) Variable Mean StDev Minimum Median Maximum Range Mode Income ($1000) 43.74 14.64 21.00 43.00 67.00 46.00 55 Size 3.420 1.739 1.000 3.000 7.000 6.000 2 Credit Balance($) 3976 932 1864 4090 5678 3814 3890 Variable Mode Skewness Kurtosis Income ($1000) 4 0.05 -1.29 Size 15 0.53 -0.72 Credit Balance($) 2 -0.14 -0.74 The statistics indicates that the mean income is 43.74; the median income is 43.00 and the mode income is 55. The standard deviation is given approximately as 1.74. Maximum income is 67,000 and the minimum income is 21,000 and a Standard Deviation of 14.64. The second individual variable is household size the mean household size of the sample is 3.42, the median household size of the sample is 3 and the mode household size of the sample is 2 and the standard deviation is 1.739. The maximum household size is 7 and the minimum household size is