Illustrative Example Sample Clauses

Illustrative Example. (i) Chemical A was dissolved in 0.010 M pH 7.0 buffer. The solution was filtered through a 0.2 )m filter, air saturated, and analyzed. It contained 1.7×10 ¥5 M A, five-fold less than its water solubility of 8.5×10 ¥5 M at 25 °C. A uv spectrum (1-cm path length) versus buffer blank showed no absorbance greater than 0.05 in the wavelength interval 290 to 825 nm, a condition required for the Phase 2 pro- tocol. The 180 mL mixture was diluted by the addition of 20 mL of SHW stock solution. (ii) The SHW solution of A was photolyzed in sealed quartz tubes (12×100 mm) in the fall season starting on October 1. At the end of 1 and 2 days, respectively, the concentration of A was found to be 1.13×10 ¥5 M and 0.92×10 ¥5 M compared to unchanged dark controls (1.53×10 ¥5 M). (iii) The tube photolysis rate con- stant of chemical A was calculated from Equation 2 under paragraph (c)(2)(vi)(B) of this section. The first time point at day 1 was used because the fraction of A remaining was in the range 20 to 80 percent:

Illustrative Example. (i) From Phase 2 testing, under paragraph (c)(6)(iii) of this section, chemical A was found to have a photolysis rate constant, (kp)SHW′ of 0.30 d¥1 in fall in round tubes at latitude 33° N. Using Table 1 under paragraph (d)(1)(vii) of this section for 30° N, the nearest decadic latitude, a fall value of ka equal to 333 d¥1 is found for PNAP. Substitution of (kp)SHW and ka into Equation 15 under paragraph (d)(2)(i) of this section gives [PYR] = 0.0242 M. This is the concentration of pyridine that gives an actinometer rate constant of 0.30 d¥1 in round tubes in fall at this latitude. (ii) The actinometer solution was made up by adding a volume of pyri- dine (1.95 mL) calculated from equation 16 under paragraph (d)(2)(ii) of this sec- tion to a 1 liter volumetric flask con- taining 1.00 mL of 1.00 × 10¥2 M PNAP in acetonitrile. The flask was filled to the mark with distilled water to give final concentrations of [PYR]=0.0242 M and [PNAP]=1.00×10¥5 M. Ten tubes of each of the following solutions were placed in the photolysis rack at 1,200 hours on day zero: (A) Chemical A (1.53×10¥5 M) in stand- ard SHW (0.010 M, pH 7 phosphate buff- er). (B) Chemical A (1.53×10¥5), in 0.010 M, pH 7 phosphate buffer. (C) SHW standard solution diluted with water 0.90 to 1.00 to match solu- tion A. (D) PNAP/PYR actinometer solution. Ten additional foil-wrapped controls of each mixture were taped to the bottom of the rack. (iii) The test chemical had been placed in category B, Table 2 under the paragraph (d)(2)(iv) of this section, on the basis of its Phase 2 rate constant under paragraph (c) of this section. Ac- cordingly, two tubes of each irradiated solution and two tubes of each blank solution were removed at 0, 1, 2, 4, and 8 days at 1,200 hours. The averaged ▇▇▇- lytical results obtained at the end of the experiment are shown in the fol- lowing Table 3. Table 3—Chemical Analytical Results for Illustrative Example, Phase 3 Day 105[C]SHW, M 105[C]W, M ASHW370 105 [PNAP], M 0 ...................................................................................... 1.53 1.53 0.0500 1.00 1 ...................................................................................... 1.03 1.40 0.0470 0.810 2 ...................................................................................... 0.760 1.30 0.0440 0.690 4 ...................................................................................... 0.300 1.01 0.0370 0.380 8 ...............................................................

Illustrative Example. Executive has been awarded 100 PSUs for Period One, 100 PSUs for Period Two and 100 PSUs for Period Three. If for Period One the Operating Income is $ (75% of the Operating Income Performance Target) and Operating Margin is (less than Threshold), for Period Two the Operating Income is $ ( % of the Operating Income Performance Target) and the Operating Margin is ( % of the Operating Margin Performance Target) and for Period Three, Operating Income is $ ( % of the Operating Income Performance Target) and Operating Margin is ( % of the Operating Margin Performance Target), then provided that the Additional Vesting Condition is met, Shares shall be delivered to the Executive between March 5 and March 15, 201 , plus a cash payment equal to shares multiplied by the Fair Market Value of a Share on March 5, 201 . The shares were determined by adding (1) Shares for Period One, (2) Shares for Period Two ( Shares for meeting the Operating Income Performance Target and Shares for meeting the Operating Margin Performance Target) and (3) ) for Period Three ( shares for achieving % of the Operating Income Performance Target and shares for achieving % of the Operating Margin Performance Target), rounding down by fractional shares.

Illustrative Example. If the minimum hourly rate is $30 and the Agreement provides for a wage increase of 3% on 1 December 2024, but the Commonwealth Government’s direction is to increase the applicable hourly rate by $3 on 1 October 2024, then, as a term of this Agreement, the Employer will increase the hourly rate to $33 on 1 October 2024 and then apply the 3% wage increase on 1 December 2024 to the hourly rate of $33.

Illustrative Example. For illustrative purposes only, below is a hypothetical example of the calculation of the respective vote fractions, assuming one vote per Underlying Share, the 90-Day VWAP of the Asset Certificate is $350.00 and the 90-Day VWAP of the Dividend Certificate is $50.00: As a result, in this illustrative example, each Asset Certificate would be allocated 0.875 of a vote per certificate, and each Dividend Certificate would be allocated 0.125 of a vote per certificate.

Illustrative Example. Hours Worked PP1 + Hours Worked PP2 / 160 x 8 = Holiday Pay 32 hrs.(PP1) + 56 hrs.(PP2) / 160 x 8 = 4.4 hrs. Holiday Pay

Illustrative Example. Inputs: S = Underlying Security Price X = Exercise Price PV(X) = Present value of the Exercise Price, discounted at a rate of R = X * (e^-(R * T)) V = Volatility R = continuously compounded risk free rate = 2 * [ ln (1 + Interest Rate / 2) ] T = Time to Expiration W = warrant value per underlying share Z = number of shares underlying warrants Value = total warrant value Formulaic inputs: D1 = [ ln [ S / X ] + (R + (V^2 / 2)) * T)] ÷ (V * ÖT) D2 = [ ln [ S / X ] + (R - (V^2 / 2)) * T)] ÷ (V * ÖT) W = [N(D1) * S] — [N(D2) * PV(X)] Where “N” is the cumulative normal probability function Value = W * Z

Illustrative Example. The Indicative Price Bids shall be submitted by the various bidders in the following format:

Illustrative Example. If actual performance is between performance levels, the Applicable Percentage will be prorated between such performance levels. The following example illustrates the method of such proration. Assume that in the year 2007 the actual Operating Ratio was 79.6%, actual EBITDA was $551 million and actual ROCE was 8.8%. The difference between the 2007 Target Operating Ratio and the actual 2007 Operating Ratio is .2, representing 15.4% of the difference between the 2007 Target Operating Ratio and the 2007 Maximum Operating Ratio (i.e., .2/(79.8-78.5)). Thus, the Operating Ratio earned percentage before weighting would be 115%. The difference between the 2007 Target EBITDA and the actual 2007 EBITDA would be $2 million, representing 2% of the difference between the 2007 Target EBITDA and the 2007 Maximum EBITDA (i.e., 2/(649-549)). Thus, the EBITDA earned percentage before weighting would be 102%. The difference between the 2007 Target ROCE and the actual 2007 ROCE would be .2, representing 13% of the difference between the 2007 Target ROCE and the 2007 Maximum ROCE (i.e., .2/(10.1-8.6)). Thus, the ROCE earned percentage before weighting would be 113%. Finally, each metric would be multiplied by the appropriate weighting factor and the weighted earned percentages would be added together to determine the earned percentage. In this example the weighted earned percentage would be 111.25%, as demonstrated in the table below: A. 2007 Actual 78.6% $551 million 8.8%