Gardenscape, Villa of Livia essays

Gardenscape, Villa of Livia essays The Gardenscape at Villa of Livia in Primaporta, Italy (Livia was the wife of Emperor Augustus) decorates a vaulted, partially subterranean chamber of the villa. The dry fresco is done in the second style of Roman art and depicts an ideal garden scene. The second style describes the artist's attempt to dissolve a room's confining walls and replace them with the illusion of an imaginary three-dimensional world. This can be seen in the artist's lack of using framing devices, thus creating a picture-window wall. Through my research of this painting I have discovered that it uses atmospheric perspective- indicating depth by the increasinly blurred appearance of objects in the distance. This is illustrated by the fence, trees and birds in the foreground, which are precisely painted and the details of dense foliage in the background, which are indistinct. Since the painting is set up horizontally, objects look wider rather than taller, and horizontal bands of fence, background and border have been created so your eye seems to see a wide-angle view. The lines of the work creat movement and a natural beauty throughtou the composition. An evident line is in the foreground. Together the trees and birds, which are curvy and flowing, creat an almost vine-like effect. The more detailed parts of the painting have darker outlines that make them stand out, creating a focal point for your eye. The whole design of the painting is based on soft, natural forms flowing together in a harmonius movement, reflecting the beauty of nature. Also, through my research, I have discovered that in the second style era of wall paintings the people of Rome wanted always to be in the beauty of nature. And so they created homes that would allow outside light and air into the house and they adorned the walls with beautiful paintings of ideal gardens. The gardens and fruits Romans produced were important to the people in cities like Primaporta. Cultivate...

Probabilities for Rolling Three Dice

Probabilities for Rolling Three Dice Dice provide great illustrations for concepts in probability. The most commonly used dice are cubes with six sides. Here, we will see how to calculate probabilities for rolling three standard dice. It is a relatively standard problem to calculate the probability of the sum obtained by rolling two dice. There are a total of 36 different rolls with two dice, with any sum from 2 to 12 possible. How does the problem change if we add more dice? Possible Outcomes and Sums Just as one die has six outcomes and two dice have 62 36 outcomes, the probability experiment of rolling three dice has 63 216 outcomes. This idea generalizes further for more dice. If we roll n dice then there are 6n outcomes. We can also consider the possible sums from rolling several dice. The smallest possible sum occurs when all of the dice are the smallest, or one each. This gives a sum of three when we are rolling three dice. The greatest number on a die is six, which means that the greatest possible sum occurs when all three dice are sixes. The sum of this situation is 18. When n dice are rolled, the least possible sum is n and the greatest possible sum is 6n. There is one possible way three dice can total 33 ways for 46 for 510 for 615 for 721 for 825 for 927 for 1027 for 1125 for 1221 for 1315 for 1410 for 156 for 163 for 171 for 18 Forming Sums As discussed above, for three dice the possible sums include every number from three to 18. The probabilities can be calculated by using counting strategies and recognizing that we are looking for ways to partition a number into exactly three whole numbers. For example, the only way to obtain a sum of three is 3 1 1 1. Since each die is independent from the others, a sum such as four can be obtained in three different ways: 1 1 21 2 12 1 1 Further counting arguments can be used to find the number of ways of forming the other sums. The partitions for each sum follow: 3 1 1 14 1 1 25 1 1 3 2 2 16 1 1 4 1 2 3 2 2 27 1 1 5 2 2 3 3 3 1 1 2 48 1 1 6 2 3 3 4 3 1 1 2 5 2 2 49 6 2 1 4 3 2 3 3 3 2 2 5 1 3 5 1 4 410 6 3 1 6 2 2 5 3 2 4 4 2 4 3 3 1 4 511 6 4 1 1 5 5 5 4 2 3 3 5 4 3 4 6 3 212 6 5 1 4 3 5 4 4 4 5 2 5 6 4 2 6 3 313 6 6 1 5 4 4 3 4 6 6 5 2 5 5 314 6 6 2 5 5 4 4 4 6 6 5 315 6 6 3 6 5 4 5 5 516 6 6 4 5 5 617 6 6 518 6 6 6 When three different numbers form the partition, such as 7 1 2 4, there are 3!  (3x2x1) different ways of permuting these numbers. So this would count toward three outcomes in the sample space. When two different numbers form the partition, then there are three different ways of permuting these numbers. Specific Probabilities We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 0.5%Probability of a sum of 4: 3/216 1.4%Probability of a sum of 5: 6/216 2.8%Probability of a sum of 6: 10/216 4.6%Probability of a sum of 7: 15/216 7.0%Probability of a sum of 8: 21/216 9.7%Probability of a sum of 9: 25/216 11.6%Probability of a sum of 10: 27/216 12.5%Probability of a sum of 11: 27/216 12.5%Probability of a sum of 12: 25/216 11.6%Probability of a sum of 13: 21/216 9.7%Probability of a sum of 14: 15/216 7.0%Probability of a sum of 15: 10/216 4.6%Probability of a sum of 16: 6/216 2.8%Probability of a sum of 17: 3/216 1.4%Probability of a sum of 18: 1/216 0.5% As can be seen, the extreme values of 3 and 18 are least probable. The sums that are exactly in the middle are the most probable. This corresponds to what was observed when two dice were rolled.