However, the CO increased with increase in the intensity of exercise with the CO for mild exercise increasing to 8,820 and that for moderate exercise increasing to 18,375. This was an increase of 117% and 353% respectively. The baseline mean arterial pressure (MAP) was 88.3 mmHg. With the introduction of mild exercise, the MAP reduced by 8.66 mmHg to 79.67 mmHg. Moderate exercise had the highest MAP at 99 mmHg, which was an increase of 10.67 mmHg from the baseline MAP.
They were however significantly different (P < 0.05) at 10-14 weeks old from the other two genotype groups. The PXP genotype similarly manifested a higher body weight gain from Table 1: Mean growth performance of three genotypes of guinea fowl Weeks Parameter Genotypes SEM P × P B × B B×P Wk2 Body weight (g) 100.79 97.62 99.91 2.40 Body weight gain (g/bird) 75.96 71.92 73.22 2.35 2-weekly feed intake (g/bird) 203.17 203.17 203.17
All the inscribed angles in each of the circles are equal (inscribed angles on the same chord law) and these sets of equal angles decrease with an increase in size of the circle (exterior angle theorem for a triangle). For example, angle AFB = angle ACB and angle ADB>AGB. Consequently, the optimal position would be found where the conversion line forms a tangent with one of the circles as it provides the largest angle along the conversion line (line TF) than any other point. This is shown in Figure 4
The bessbug was able to carry 6 weights which has the mass of 18.85 grams, while ten times the mass of a bessbug is 17.6 grams. The maximum amount of weights the bessbug was able to pull was 10 weights which is equal to 31.46 grams. The data shows the strength of the bessbug and how much the bessbug is able to carry in all. The data occurred because the capacity of the bessbug allows the bessbug to carry 10 weights. The data supports the hypothesis because the bessbug was able to lift ten times the weight of the bessbug which is 17.6 grams, and much more because the total amount the bessbug was able to pull was 31.46 grams.
For the 16PF primary scales, test–retest reliabilities average 0.80 over a two-week interval (ranging from 0.69 to 0.87), and 0.70 over a two-month interval (ranging from 0.56 to 0.79). The five global scales of the 16PF Questionnaire show even higher test–retest reliabilities (they have more items); they average 0.87 for a two week interval (ranging from 0.84 to 0.91), and 0.78 for a two-month interval (ranging from 0.70 to 0.82). One important source of validity for the 16PF Questionnaire has been factor-analytic studies of the structure of the primary and global traits across diverse samples of people. These studies have used exploratory and confirmatory factor analysis to confirm the number, identity, and independence of the primary factors; and to confirm the number, identity, and primary factor make-up of the global factors. There are numerous applications for 16PF and it is useful in the following settings: • Career Choices • Counselling and correctional
This table presents the information in colony factoring units per gram, or CFU/G. To obtain this number, a series of calculations were conducted. Beginning with the total number of colonies in each dilution, this number was divided by 0.1. Then the number found from the division explained previously was then multiplied by 10 to the power of n, where n was the number in the dilution set. For example: dilution 4 would solved using the answer to the division problem and multiplying by 10 to the 4th power.
The highest growth and red pigment production obtained from medium containing amino acid was L- tryptophan (4.75 ± 0.09 A500) and maximum dry cell mass was (3.9 ± 0.02 g/l) as shown in fig. 9. Juzlova et al (1996) described that amino acid addition in growth medium of Monascus lead to increase red pigment production than other pigment. Metal ions additions showed a great effect on Monascus both growth and red pigment production (Fig. 10).
If Portugal has to spent the labor of 80 people for one year on the production of the wine or 90 workers on the cloth, England may need 120 and 100 people for this. Although, Portugal has an absolute advantage, which means it requires less resources on both goods, it is better to produce the wine only and exchange it to the cloth from England, since England has comparative advantage in the cloth: if it uses 100 labor for the cloth, it loses 5/6
In this example, two dies are open to differ whereas the third die is not. For that reason, degree of freedom is equal to 2. For data with one category: Degree of freedom = number of observations -1 For data with more than one category represented in a table: Degrees of freedom = (number of rows in the table) – 1 X (number of columns in the table) -1 The chi-square test is used to determine whether two variables are independent or not. If two variables are dependent on each other, their values have a tendency to progress together, either in the opposite direction or in the same. Example Consider a data set of 100 individuals divided into categories of Male, Female and university admission (Yes/No).
Based on making 8 of every 33 years leap years, it was more accurate than the present Gregorian calendar, and it was adopted in 1075 by Malik-Shāh. Khayyam found that 1,029,983 days made 2,820 years. This gives a tropical year length of 365.2422 days to seven significant figures. Although it has become fashionable to quote more decimal places than this, Khayyam’s input of 1,029,983 days contains seven significant figures, so it is unreasonable to quote more than this number of significant figures in the calculated year length. Today we know that the length of a tropical year actually changes by as much as 30 minutes from year to year.