O’Sullivan’s arguments for annexation. How does he frame his support.pdf

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Briefly explain four functions of stems. Solution Function of Stems: 1. It helps store water , as in cacti, and the products of photosynthesis, as in the trunk of sago palm (Metroxylon sagu) and sweet palm (Argenga pinnata) which store large stock of starch 2. Young green stem also performs a minor role in the production of food through the process of photosynthesis, but in some species (e.g. cactus) the stem is the chief photosynthesizing organ 3. It supports the leaves, flowers and fruits and connect them with roots. In trees and shrubs the main stem or trunk provides a strong columnar structure from which branches are attached raising the leaves upward to be expressed more fully to the sun 4. The plant stem serves as a means of asexual reproduction in many species..

Briefly explain four functions of stems. SolutionFunction of S.pdf

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AVR was one of the first microcontroller families to use A. on-chip flash memory for program storage B. one-time programmable ROM, EPROM, or EEPROM C. both A and B EMBEDDED SYSTEMS is a computer system with a dedicated function within a larger A. digital and logical system B. mechanical or electrical system C. both A and B An integrated circuit that contains all the functions of a central processing unit of a computer is A. Microprocessor B. Microcontroller C. Both A and b A main system bus, ram, rom/flash, and typically a set of other devices such as a programmable interrupt controller, analog/digital converters, any number of general purpose 10 devices A. Microcontroller B. Microprocessor C. None of the above ASSEMBLY LANGUAGE is an assembly (or assembler) language, often abbreviated asm, is a A. low-level programming language for a computer B. high-level programming language for a computer A. C. both A and B Solution AVR was one of the first micro controller families to use on chipflash memory for program storage Embedded syatems are mechanical or electrical syatems An integrated circuit that has all functions of a cpu is micro processor Assembler language is low level programming language for a computer A main system........... Micro processor.

AVR was one of the first microcontroller families to use A. on-chip .pdf

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Complete the implementation of the binary search tree started in class. You know you are done when the main function runs correctly with all commented lines uncommented, and printing shows the expected behavior of a BST. CODE: # TODO: implement inorder # TODO: implement find # TODO: implement maximum, minimum # TODO: implement successor, predecessor # TODO: implement delete def main(): bst = Node(3, None) for k in [5, 7, 4, 5, 6]: bst.insert(k) # print(bst.inorder()) # for k in [4, 3, 6, 2, 1, 7]: # print(\"find {}: {}\".format(k, bst.find(k) != None)) # for k in [4,3, 2]: # n = bst.find(k) # if n != None: # s = n.successor() # print(\"successor of {}: {}\".format(k, s.key if s != None else None)) # s = n.predecessor() # print(\"predecessor of {}: {}\".format(k, s.key if s != None else None)) # s = n.delete() # print(\"deleted {}\".format(s.key)) # print(bst.inorder()) class Node: def __init__(self, key, parent): self.key = key # self is Python for \"this\" self.parent = parent self.left = None self.right = None def inorder(self): \'\'\'returns list of the keys in an inorder traversal of the tree rooted at self\'\'\' # TODO: replace \"pass\" keyword with actual code # Hint: li.append(value) appends a value to list li # Hint: li = list1 + list2 returns the concatnenation of list1 and list2 # Hint: use dot notation to call method: inorder(node) ---> node.inorder() # To test: uncomment line 10 and click \"Run\" li = [] # empty list if self.left is not None: pass pass if self.right is not None: pass return li def insert(self, key): \'\'\'inserts a new node with given key into subtree rooted at self node\'\'\' if key < self.key: if self.left is None: # None is Python for null, \"is\" = \"==\", \"is not\" = \"!=\" self.left = Node(key, self) else: self.left.insert(key) else: if self.right is None: self.right = Node(key, self) else: self.right.insert(key) def find(self, key): \'\'\'returns a node in the subtree rooted at self with this key or None if no such key exists\'\'\' # Hint: similar to insert() # Test: uncomment lines 11 and 12, click \"Run\" pass def minimum(self): \'\'\'returns the node with minimum key in the subtree rooted at self\'\'\' pass def maximum(self): \'\'\'returns the node with maximum key in the is subtree rooted at self\'\'\' pass def successor(self): \'\'\'returns successor of the self node or None if it does not have one\'\'\' # TODO: fill the if statement if self.right is not None: pass c, p = self, self.parent while p is not None and c is p.right: c, p = p, p.parent return p def predecessor(self): \'\'\'returns the predecessor of the self node or None if it does not have one\'\'\' pass def delete(self): \'\'\'deletes the self node from the tree and returns that node\'\'\' parent, left, right = self.parent, self.left, self.right # case 0: no children if left is None and right is None: if parent is not None: if self is parent.left: parent.left = None else: parent.right = None return self # case 1: one or two children, no pa.

Complete the implementation of the binary search tree started in cla.pdf

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A researcher was interested in studying Americans email habits. She suspected that Americans spend less than 7 hours a week answering their email.The General Social Survey in 2004 included a question that asked about the number of hours that the respondent spend on email per week. The General Social Survey in 2002 asked 1,264 respondents this question. The sample mean number of hours was 6.02 and the sample standard deviation was 7.80. Find the test statistic.A. 4.47 Solution Test Statistic Population Mean(U)=7 Sample X(Mean)=6.02 Standard Deviation(S.D)=7.8 Number (n)=1264 we use Test Statistic (t) = x-U/(s.d/Sqrt(n)) to =6.02-7/(7.8/Sqrt(1263)) to =-4.47 ANS:B.-4.47.

A researcher was interested in studying Americans email habits. She .pdf

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A 50 mL stock sample of cells is diluted 1/10 six times. If 0.1 mL of the most dilute sample generates 75 colonies on an agar plate, calculate the original concentration of bacteria present in the original sample? Solution Answer: Given: Dilution factor= 1/10^6 Number of colonies= 75 Volume of sample = 0.1 mL to the plate CFU/ml = ? cfu/ml = (no. of colonies)/ dilution factor x volume of culture plate cfu/ml = (75)/ (1/10^6 x 0.1) cfu/ml = 75 x 10^7.

A 50 mL stock sample of cells is diluted 110 six times. If 0.1 mL o.pdf

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Why are clocks (or timers) discussed in the context of IO systems.pdf

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why the SSB receiver re-injects the missing carrier? Why add the carrier back in after downshifting the incoming modulated signal instead of before why the SSB receiver re-injects the missing carrier? Why add the carrier back in after downshifting the incoming modulated signal instead of before Solution Re-injection of carrier is done by BFO,this re-injection of carrier allows easy de-modulation of RF and even IF signal also.if injected carrier is at the wrong frequencyrelative to the station then the audiio will become disorted..

why the SSB receiver re-injects the missing carrier Why add the c.pdf

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80 µl of 10^-5 dilution of E. coli culture was plated on LB agar and incubated at 37 degree Celsius overnight. Next day, you observed 212 colonies on this plate. Based on this information, calculate the viable count (cfu/mL) of the original E. coli culture. Solution Colony forming units (CFU)/mL (from plate C) = (no. of colonies x dilution factor) / volume of culture plate Number of colonies observed is = 212 Dilution factor is = 10^5 Volume of culture plate =80 µl Colony forming units (CFU)/mL (from plate C) = 212*10^5/ 0.08 ml = 26.50* 10^7 CFU/ml.

80 µl of 10^-5 dilution of E. coli culture was plated on LB agar and.pdf

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Which of the following statements about pyruvate is most accurate? Pyruvate can be produced only from carbohydrates; Pyruvate dehydrogenase complex does not belong to the Krebs cycle. Therefore, pyruvate dehydrogenase is not a step-controlling enzyme for the Krebs cycle; Pyruvate can be reversibly converted to phosphoenolpyruvate by pyruvate kinase; Pyruvate is a main precursor of oxaloacetate in the Krebs cycle; Pyruvate dehydrogenase converts pyruvate to acetyl CoA and generates one NADH that is equivalent to 2ATP if the NADH is transferred to mitochondria via glycerol 3-phosphate shuttle system. Solution Correct answer is B Pyruvate dehydrogenase complex does not belong to Krebs Cycle.Therefore pyruvate dehydrogenase is not a step controlling enzyme for krebs cycle. PDH complex catalyses the conversion of Pyruvate into Acetyl CoA. Acetyl CoA then enters Krebs cycle and forms oxaloacetate. Therefore action of PDH comples does not determine the control of krebs cycle,..

Which of the following statements about pyruvate is most accurate P.pdf

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Which CLI modes let you use the show running-config command? A. User B. Privileged C. Global Configuration D. Setup E. All of the above Solution B is correct. You can run the show running-config command in Privileged (Enable) modes. A, C, D, and E are incorrect. A, C, and D are incorrect because the show running- configcommand is not available in these modes. E is incorrect because choices A, C, and D are incorrect..

Which CLI modes let you use the show running-config commandA. Use.pdf

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What is meant by \"star activity\" with respect to a restriction enzyme. I just need a 1-2 sentence explanation Solution 1. Star activity is a general property of Restriction endonuclease. 2. During some non-standard reaction conitions, specific restriction enzymes are can cleave the sequences which are similar , but not identical ,to their efined recognition seqences. 3.Such kind of soecificity has been calle \" Star activity\"..

What is meant by star activity with respect to a restriction enz.pdf

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Type III hypersensitivities are also referred to as immune complex mediated hypersensitivities. Which cell is responsible for clearing the antibody-antigen complexes that are formed? In type I hypersensitivity, prior sensitization to the antigen/allergen is not required because it\'s an immediate type of reaction. True False Which cell types participate in antibody mediated attacks of either cells or red blood cells in type II hypersensitivities? Solution Hypersensitive is the group of unfavourable reactions produced by the normal Immune system. It includes allergies and Autoimmunity. Type-III immunity is immune complex mediated diseases that include Serum sickness, Rheumatoid arthritis etc. The mediators are IgG, complement and Neutrophills. 7. Answer is Macrophages. 8. False: Prior sensitization with allergin, leads to IgE production, then they bind to mast cells. Again encounter wits same allergin, bind to these mast cell bound IgE. Finally leads to degranulation and Type-1 hypersensitivity. 9. All three types of cells: Macrophages, NK cells and Neutrophils.

Type III hypersensitivities are also referred to as immune complex me.pdf

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True, False, Uncertain with Explanation: (a) Including an interaction term between two independent variables, X1 and X2, allows for the measurement of the effect of a unit increase in X1 and X2, above and beyond the sum of the individual effects of a unit increase in the two variables alone. Solution When an interaction is present, the lines in a graph of the means will not be parallel. Another way of describing the interaction is to say that the effects of one factor are not the same at all levels of the other factor.Also we woll say that “the simple effects are different,” because a simple effect looks at the effects of one factor at each level of the other factor. An interaction effect is a change in the simple main effect of one variable over levels of the second. An A X B or A BY B interaction is a change in the simple main effect of B over levels of A or the change in the simple main effect of A over levels of B.In either case the cell means cannot be modeled simply by knowing the size of the main effects. An additional set of parameters must be used to explain the differences between the cell means. These parameters are collectively called an interaction. If interaction is significant, determine whether interactions are important. If not, can examine main effects If interaction present & important, determine whether interaction is simple or complex. . For simple interactions, can still talk about the main effects of A at each level of B For complex interaction, must simply consider all pairs of levels as separate treatments. If interaction effects are very small compared to main effects or only apparent in a small number of treatments, then they are probably unimportant. We can proceed by keeping interaction in the model, but using marginal means for each significant main effect individually The interaction effect is so large and/or pervasive that main effects cannot be interpreted on their own. This is TRUE statement..

True, False, Uncertain with Explanation(a) Including an interacti.pdf

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Traits that are detrimental to the long-term survival of an individual can occur in a population (select all that apply): a. if they arise due to a mutation. b. Never c. when the trait increases an individual\'s chance of reproducing. d. because there has to be variation in the population. e. if they enter the population through gene flow from another habitat where the traits are favorable. Solution a. if they arise due to a mutation. d. because there has to be variation in the population..

Traits that are detrimental to the long-term survival of an individu.pdf

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The sampling distribution of the sample mean is the probability dist.pdf

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The lantana has which inflorescence morphology type campanulte tub.pdf

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The intensity of the sound of traffic at a busy intersection was measured at 2.0 × 101 W/m2. Find the decibel level. (Round your answer to the nearest whole number.) Solution Sound intensity of traffic at a busy intersection is given by I = 2×10^1 W/m². Now, Reference sound intensity Io = 10^12 W/m² (Threshold of hearing) Reference sound intensity level LIo = 0 dB (Threshold of hearing level) we need to find sound intensity level LI in dB when entering sound intensity I = 2×10^1 W/m². Sound intensity level LI = 10×log (I / Io) dB = 10×log (2×10^1 / 10^12)=10*11.30 dB = 113.01 dB =113dB on round off.

The intensity of the sound of traffic at a busy intersection was mea.pdf

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The garden pea is described as \"self-fertilizing.\" What does this mean and why was this significant to the success of Mendel\'s inheritance experiments? Solution The pea plant which Mendel chose for conducting experiments, is most ideal for controlled breeding, since it can easily be subjected to cross pollination as well as self pollination. This gives the success to Mendel\'s inheritance experiments. The main other resons for mendal success are.

The garden pea is described as self-fertilizing. What does this .pdf

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The equation for the time of one swing of a pendulum is given by T = 2 pi Squareroot L/32 where T is the time in seconds, pi almostequalto 3.14, & L is the length in feet. What is the length of the pendulum if it makes one (1) swing in 4 seconds, according to the formula? Solution T=2*pi*(L/32)^.5 Squaring on both sides (T/(2*pi))^2=L/32 Given time =4 sec L/32= (4/(2pi))^2 L/32= 4/10 L=128/10 L=12.8 Length of pendulum =12.8 units L/32=.

The equation for the time of one swing of a pendulum is given by T = .pdf

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The assumption that arrivals follow a Poisson probability distributi.pdf

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